Python 3.8 (pre-release), 46 47 48 bytes

-1 thanks to Neil

-1 thanks to ceilingcat

Takes quite a long time for inputs like 2**31-1...

f=lambda n:n*4in[(x*~x)**2for x in range(n+1)]

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Explanation for original 48-byte version

f=lambda n:n in[(x*x+x)**2/4for x in range(n+1)]

The xth triangular number is (x*x+x)/2, and we're looking for the square of that: ((x*x+x)/2)**2

We can save two characters by squaring the numerator and denominator separately: (x*x+x)**2/4

We need range(n+1) instead of range(n) to return True when n is 1

AWK, 28 25 bytes

$0=(1+8*$0^.5)^.5%2==1{}1

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Explanation: uses the formula in Eonema's answer. $0 is reassigned in the pattern, so it's printed automatically if it's 1; otherwise, we test for the pattern /0/ and print that.

Saved three bytes thanks to Marius_Couet.

Previous solution: 42 41 bytes

{for(b=0;$0>b;b+=a++**3);print($0~b);a=0}

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Explanation:

• The empty for-loop repeatedly adds successive cubes a^3 to b until b is at least as large as the input line.
• Print 1 when $0 is exactly b, or 0 otherwise (the regex match works because b>=$0).
• Reset a for the next line.

JavaScript (V8), 22 bytes

n=>(8*n**.5+1)**.5%1>0

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This is Arnauld's code with -1 by Shaggy, I'm just the messenger. Posting because of lack of a JavaScript answer.

Uiua, 9 8 bytes

∊\+⇡999√

Pad

-1 byte: Tbw - switched from checking \$n\in \sum k^3\$ to \$\sqrt{n}\in \sum k\$

Creates a list of \$\sum_{k=0}^n k \ (0\le n\le 999)\$, then checks membership of \$\sqrt{n}\$. Since the largest squared triangular number under \$2^{31}-1\$ is the 303^rd, this range must be 3 digits so we go up to the 999^th.

Wolfram Language (Mathematica), 18 bytes 21 bytes

OddQ@√(1+8√#)&

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Edit

-3 thanks to att to shorten Sqrt

Google Sheets, 26 bytes 76 bytes

=0=mod(sqrt(1+8*A1^0.5),1)

Expects the integer in cell A1. Returns true or false.

See A000537. Thanks to Arnauld.

x86-64 machine code, 17 bytes

31 C9 89 C8 F7 E1 F7 E1 FF C1 29 C7 77 F4 19 C0 C3

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Following the standard calling convention for Unix-like systems (from the System V AMD64 ABI), this takes a 32-bit integer in EDI and returns a 32-bit integer in EAX, which is 0 if the number is in the sequence and -1 if it is not.

In assembly:

f:  xor ecx, ecx    # Set ECX to 0.
r:  mov eax, ecx    # Set EAX to ECX.
    mul ecx         # Multiply EAX by ECX, producing the square.
    mul ecx         # Multiply EAX by ECX, producing the cube.
    inc ecx         # Add 1 to ECX.
    sub edi, eax    # Subtract the cube from the input number.
    ja r            # Jump back, to repeat, if it stays positive.
    sbb eax, eax    # Subtract EAX+CF from EAX, making it -CF.
                    #  CF was set by the previous SUB instruction;
                    #  it's 1 iff the value overflowed to negative.
    ret             # Return.

Charcoal, 13 bytes

№Ｅ⊕θＸ×ι⊕ι²×Ｎ⁴

Try it online! Link is to verbose version of code. Outputs a Charcoal boolean, i.e. - for a sum of cubes, nothing if not. Explanation: Port of my golf to @V_R's Python answer.

   θ            Input `n` as a string
  ⊕             Cast to integer and increment
 Ｅ              Map over implicit range
      ι         Current value
     ×          Multiplied by
        ι       Current value
       ⊕        Incremented
    Ｘ           Raised to power
         ²      Literal integer `2`
№               Contains
           Ｎ    Input `n` as a number
          ×     Multiplied by
            ⁴   Literal integer `4`
                Implicitly print

If potential floating-point accuracy is acceptable, then for 10 bytes:

⁼¹﹪₂⊕×⁸₂Ｎ²

Try it online! Link is to verbose version of code. Explanation: Port of @Eonema's R answer.

T-SQL (set-based), 161 bytes

DECLARE @ INT=9 SELECT COUNT(*) FROM (SELECT POWER(n*(n+1)/2,2)s FROM (SELECT TOP 999 ROW_NUMBER()OVER(ORDER BY(SELECT 1))n FROM master..spt_values)x)y WHERE s=@

T-SQL (iterative), 121 bytes

DECLARE @x INT=9DECLARE @ INT=1,@s INT=0 a:SET @s+=@*@*@ IF @s<@x AND @<303 BEGIN SET @+=1 GOTO a END PRINT 1-SIGN(@s^@x)

R, 23 bytes

\(x)(1+8*x^.5)^.5%%2==1

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Explanation:

The n^th square triangle number is \$x = (\frac{n(n+1)}{2})^2\$, so \$2\sqrt{x}=n^2+n\$, giving \$n=\frac{-1+\sqrt{1+8\sqrt{x}}}{2}\$. Since \$n\$ has to be an integer, that means \$\sqrt{1+8\sqrt{x}}\$ has to be an odd integer, i.e., \${\sqrt{1+8\sqrt{x}}}\mod{2}=1\$.

Ruby, 31 bytes

->n{(1..n).any?{|x|0==n-=x**3}}

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Brachylog, 8 bytes

≥ℕ⟦^₃ᵐ+?

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Explanation

≥ℕ          Get an integer in [0, input].
  ⟦         Get the range [0, that integer].
   ^₃ᵐ      Map cube over that range.
      +     Get its sum.
       ?    Assert that sum equals the input.

APL+WIN, 12 bytes

Prompts for integer.

⎕∊+\(⍳303)*3

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Japt, 7 bytes

õ³å+ øU

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õ³å+ øU     :Implicit input of integer U
õ           :Range [1,U]
 ³          :Cube each
  å+        :Cumulative sums
     øU     :Contains U?

Retina 0.8.2, 45 bytes

.+
$*
((^1|1\2)+)(?<=(1)*1)(?<-3>\1)*$(?(3)^)

Try it online! Link includes some test cases. Explanation: Converts the input to unary, then searches for a triangular number whose square is the input.