TITLE: Proving that a list of perfect square numbers is complete
QUESTION [8 upvotes]: Well, I have a number  $n$ that is given by:
$$n=1+12x^2\left(1+x\right)\tag1$$
I want to find $x\in\mathbb{Z}$ such that $n$ is a perfect square.
I found the following solutions:
$$\left(x,n\right)=\left\{\left(-1,1^2\right),\left(0,1^2\right),\left(1,5^2\right),\left(4,31^2\right),\left(6,55^2\right)\right\}\tag2$$

Is there a way to prove that this a complete set of solutions? So I mean that the solutions given in formula $(2)$ are the only ones?


My work:

We know that:
$$
1 + 12x^2 \left(1+x \right) \ge 0
  \space \Longleftrightarrow \space
  x \ge -\frac{1+2^{-2/3}+2^{2/3}}{3}
  \approx -1.07245
\tag3
$$
So we know that for $x<-1$ there are definitely no solutions.

REPLY [3 votes]: $y^2=1+12x^2(1+x) \implies  (12 y)^2 = (12 x)^3 + 12 (12 x)^2 + 144$
Magma code for positive $y$ only:
S:= IntegralPoints(EllipticCurve([0,12,0,0,144]));
for s in S do
  x:= s[1]/12;
  if x eq Floor(x) then
    print "(",x,", ",Abs(s[2]/12),")";
  end if;
end for;

Output:
( -1 ,  1 )
( 0 ,  1 )
( 1 ,  5 )
( 4 ,  31 )
( 6 ,  55 )