APL(NARS), 146 chars

r←g w;s;l;i;j;m
s←-l←⌈/3+⌊√∣(3○w),11○w⋄r←∅⋄i←l⋄→5
j←l
→4×⍳(1≥∣m)∨w=m←<i j⋄→4×⍳1≥∣w÷m⋄→4×⍳0≠m∣w⋄r←m⋄→0
→3×⍳s<j-←1
→2×⍳s<i-←1

q←{∅=k←g⍵:⍵⋄(∇⍵÷k),∇k}

//16+34+4+48+11+11+22=146

The function make answers to exercise is the recursive function q that has as input one complex number (written as aJb as a+ib) and return a list of complex numbers.

Here I suppose that the factors aJb have both a and b less than 3+⌊√∣(3○w),11○w.

It seems ok in the few test... I'm not sure if each factor is prime. Test:

  q 2
1J¯1 1J1 
  q 0J3
0J3 
  q 256
¯1J¯1 1J1 1J¯1 1J1 1J¯1 1J1 1J1 1J¯1 1J1 1J¯1 1J1 1J1 1J¯1 1J1 1J¯1 1J1 
  ×/q 256
256J0 
  q 7J9
1J2 1J¯1 3J2 
  ×/q 7J9
7J9 
  q 27J15
7J¯2 1J1 3J0 
  q 6840J585
1J¯6 1J2 3J0 2J1 3J0 1J¯6 1J4 
  ×/q 6840J585
6840J585 

It seems that each result Number in the list Is a Gaussian prime, without apply the definition of Gaussian prime.

Python 3.12+, 690 687 577 553 538 bytes

^{-003 bytes
-110 bytes thanks to @ceilingcat and @CrSb0001
-024 bytes thanks to @ceilingcat and @CrSb0001
-015 bytes thanks to @ceilingcat and @CrSb0001}

A=abs
R=lambda z:round(z.real)+round(z.imag)*1j
def G(a,b):
 if A(a)<A(b):a,b=b,a
 return G(b,r)if(r:=a-R(a/b)*b)else b
def C(n):
 a=[];N=int(A(n)**2);s=[1>0]*N;s[0]=s[1]=F=1>1;s[4::2]=[F]*(N-3>>1)
 for i in range(3,-~int(N**.5)):
  if s[i]:s[i*i::2*i]=(~-N+i*(2-i))//2//i*[F]
 for p in[i for i,v in enumerate(s)if v]:
  while N%p<1:N//=p;a+=p,
 F,P,*z=a,1
 while[]<F:
  if(a:=F.pop(0))%4>2:u=a;F.pop(0)
  else:
   while P<a!=-~pow(P:=P+1,~-a//2,a):1
   if (n-(u:=G(a,pow(P,~-a//4,a)+1j))*R(n/u)):u-=u.imag*2j
  n/=u;z+=R(u),
 return z[:-1]+[z[-1]*R(n)]

Admittedly likely my worst Python golf, it could definitely be improved. At least it passes all of the test-cases with no problem.

Also I am very glad that $$3\cdot3\cdot(2+i)(-1-2i)(4-i)(6+i)(-1+6i)=6840+585i$$and that all factors mentioned above are all prime in \$\mathbb Z[i]\$, I would not have enjoyed bug-fixing that.

The main method I used to factorize the Gaussian integers was to try to look for prime factors in the rectangle$$((I+Ji),(-I+Ji),(-I-Ji),(I-Ji))$$ where \$I,J=|z.\text{real}|,|z.\text{imag}|\$.

Original try it online! Link contains the main test-cases I was most worried about failing.

Try it online!

Jelly, 61 55 50 bytes

ÆiḤp/-,1p`¤×€×1,ıFs2S€⁸÷ÆiḞƑ$ƇỊÐḟ1;Ṫð,÷@\ḟ1
Ç€F$ÐL

Try it online! (Header and Footer formats the output)

-6 bytes thanks to @EricTheOutgolfer

-5 bytes thanks to @caird coinheringaahing

How it Works

ÆiḤp/-,1p`¤×€×1,ıFs2S€⁸÷ÆiḞƑ$ƇỊÐḟ1;Ṫð,÷@\ḟ1 - helper: outputs a factor pair of the input
ÆiḤp/                    - creates a list of possible factors a+bi, a,b>=0
      -,1p`¤×€           - extend to the other three quadrants 
              ×1,ıFs2S€  - convert to  actual complex numbers 
⁸÷                       - get quotient with input complex number
  ÆiḞƑ$Ƈ                    - keep only Gaussian numbers (those unchanged when complex parts are floored)
     ỊÐḟ                 - remove units (i,-i,1,-1)
        1;               - append a 1 to deal with primes having no non-unit factors
          Ṫð,÷@\         - convert to a factor pair
                ḟ1       - remove 1s
Ç€F$ÐL
Ç€      - factor each number
   $    - and
  F     - flatten the list
    ÐL  - until factoring each number and flattening does not change the list

Pyth, 54 51 45 42 36 bytes

 .W>H1cZ
h+.aDf!%cZT1>#1.jM^s_BM.aZ2

Try it online!

Accepts input in the form 1+2j - purely real or imaginary numbers can omit the other component (e.g. 9, 2j). Output is a newline-separated list of complex numbers, in the form (1+2j), with purely imaginary numbers omitting the real part.

This uses simple trail division, generating all gaussian integers with magnitude greater than 1 and less than the current value, plus the value itself. These are filtered to keep those that are a factor of the value, and the smallest by magnitude is chosen as the next prime factor. This is output, and the value is divided by it to produce the value for the next iteration.

Also, Pyth beating Jelly 😲 (I don't expect it to last though)

 .W>H1cZ¶h+.aDf!%cZT1>#1.jM^s_BM.aZ2ZQ   Implicit: Q=eval(input())
                                         Newline replaced with ¶, trailing ZQ inferred
 .W                                  Q   While <condition>, execute <inner>, with starting value Q
   >H1                                   Condition function, input H
   >H1                                     Is magnitude of H > 1?
                                           This ensures loop continues until H is a unit, i.e. 1, -1, j, or -j)
      cZ¶h+.aDf!%cZT1>#1.jM^s_BM.aZ2Z    Inner function, input Z
                                .aZ        Take magnitude of Z

                             _BM           Pair each number in 0-indexed range with its negation
                            s              Flatten
                           ^       2       Cartesian product of the above with itself
                        .jM                Convert each pair to a complex number
                      #                    Filter the above to keep those element where...
                     > 1                   ... the magnitude is greater than 1 (removes units)
              f                            Filter the above, as T, to keep where:
                 cZT                         Divide Z by T
                %   1                        Mod real and imaginary parts by 1 separately
                                             If result of division is a gaussian integer, the mod will give (0+0j)
               !                             Logical NOT - maps (0+0j) to true, all else to false
                                           Result of filter are those gaussian integers which evenly divide Z
           .aD                             Sort the above by their magnitudes
          +                         Z      Append Z - if Z is ±1±1j, the filtered list will be empty
         h                                 Take first element, i.e. smallest factor
        ¶                                  Print with a newline
      cZ                                   Divide Z by that factor - this is new input for next iteration
                                         Output of the while loop is always 1 (or -1, j, or -j) - leading space suppesses output

Perl 6, 141 124 bytes

Thanks to Jo King for -17 bytes

sub f($_){{$!=0+|sqrt .abs²-$^a²;{($!=$_/my \w=$^b+$a*i)==$!.floor&&.abs>w.abs>1>return f w&$!}for -$!..$!}for ^.abs;.say}

Try it online!

Rust - 212 bytes

use num::complex::Complex as C;fn f(a:&mut Vec<C<i64>>){for _ in 0..2{for x in -999..0{for y in 1..999{for i in 0..a.len(){let b=C::new(x,y);if(a[i]%b).norm_sqr()==0&&(a[i]/b).norm_sqr()>1{a[i]/=b;a.push(b)}}}}}}

I'm not 100% sure if this works 100% correct, but it seems to be correct for a large range of tests. This isn't smaller than Jelly, but at least it is smaller than the interpreted languages (so far). It also seems to be faster and can work through inputs of a billion magnitude in less than a second. For example 1234567890+3141592650i factors as (-9487+7990i)(-1+-1i)(-395+336i)(2+-1i)(1+1i)(3+0i)(3+0i)(4+1i)(-1+1i)(-1+2i), (click here to test on wolfram alpha)

This started out as the same idea as naive factoring of integers, to go through each number below the integer in question, see if it divides, repeat until done. Then, inspired by other answers, it morphed... it repeatedly factors items in a vector. It does this a good number of times, but not 'until' anything. The number of iterations has been chosen to cover a good chunk of reasonable inputs.

It still uses "(a mod b) == 0" to test whether one integer divides another (for Gaussians, we use builtin Rust gaussian modulo, and consider "0" as norm == 0), however the check for 'norm(a/b) != 1' prevents dividing "too much", basically allowing the resulting vector to be filled with only primes, but not taking any element of the vector down to unity (0-i,0+i,-1+0i,1+0i) (which is prohibited by the question).

The for-loop limits were found through experiment. y goes from 1 up to prevent divide-by-zero panics, and x can go from -999 to 0 thanks to the mirroring of Gaussians over the quadrants (I think?). As regarding the limitations, the original question did not indicate a valid range of input/output, so a "reasonable input size" is assumed... (Edit... however i am not exactly sure how to calculate at which number this will begin to "fail", i imagine there are Gaussian integers that are not divisible by anything below 999 but are still surprisingly small to me)

Try the somewhat ungolfed version on play.rust-lang.org

Python 2, 250 239 223 215 bytes

e,i,w=complex,int,abs
def f(*Z):
 if Z:
	z=Z[0];q=i(w(z));Q=4*q*q
	while Q>0:
 	 a=Q/q-q;b=Q%q-q;x=e(a,b)
 	 if w(x)>1:
		y=z/x
		if w(y)>1 and y==e(i(y.real),i(y.imag)):f(x,y);z=Q=0
 	 Q-=1
	if z:print z
	f(*Z[1:])

Try it online!

• -11 bytes when using Multiple Function Arguments
• -2²*² bytes when using one variable to parse couples (a,b)
• -2³ bytes when mixing tabs and spaces: thanks to ovs

Some explanation recursively decompose a complex to two complexes until no decomposition is possible...

Ruby, 258 256 249 246+8 = 264 257 254 bytes

Uses the -rprime flag.

Geez, what a mess.

Uses this algorithm from stackoverflow.

->c{m=->x,y{x-y*eval("%d+%di"%(x/y).rect)};a=c.abs2.prime_division.flat_map{|b,e|b%4<2?(1..e).map{k=(2..d=b).find{|n|n**(~-b/2)%b==b-1}**(~-b/4)%b+1i;d,k=k,m[d,k]while k!=0;c/=d=m[c,d]==0?d:d.conj;d}:(c/=b<3?(b=1+1i)**e:b**e/=2;[b]*e)};a[0]*=c;a}

Try it online!