Proving that a list of perfect square numbers is complete 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.

 A: $y^2=1+12x^2(1+x) \implies  (12 y)^2 = (12 x)^3 + 12 (12 x)^2 + 144$
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 )

A: COMMENT:
n is odd, let $n=(2m+1)^2$ ; m∈Z, then:
$4m(m+1)=12x^2(1+x)$ ⇒ $m(m+1)=3 x^2 (1+x)$
Suppose $m=3t$, then:
$t(3t+1)=x^2(1+x)$
⇒ $n=(2m+1)^2=(6t+1)^2$ 
For example $n=31^2= (6\times 5 +1)^2$ or $n=55^2=(6\times 9+1)^2$
Therefore there may be more solutions. Now let $m+1=3t$ then:
$(3t-1)3t=3x^2(1+x)$ ⇒ $(3t-1) t=x^2(1+x)$ ⇒ $n=(2m+1)^2= (6t-1)^2$
For example $n=5^2=( 6\times 1 -1)^2$
In this case there may also be some solutions.
Moreover n can be the square of negative numbers, so it can have general forms $n=[±(6t+1)]^2$ or $n=[±(6t-1)]^2$.We have to show whether equation $x^3+x^2=3t^2±t$ can have infinite solutions or limited number of solutions in Z.
