Proving a statement about perfect squares The statement I want to prove is the following:

$9+108x^2(1+x)$ is never a perfect square besides $x=0\space\wedge\space x=\pm1\space\wedge\space x=4\space\wedge\space x=6$.


My Work
Let's name the number: $\text{n}=9+108x^2(1+x)$. Because it must be a perfect square we can write: $\text{n}^2=9+108x^2(1+x)$ (where $\text{n}\in\mathbb{Z}$). Let's divide both sides by $9$: $\frac{\text{n}^2}{9}=1+12x^2(1+x)$. Now, $\frac{\text{n}^2}{9}=\left(\frac{\text{n}}{3}\right)^2$ is just another integer and perfect square. So we can write $\left(\frac{\text{n}}{3}\right)^2=\text{p}^2$ (where $\text{p}\in\mathbb{Z}$). So we have:
$$\text{p}^2=1+12x^2(1+x)\space\Longleftrightarrow\space\text{p}^2-1=12x^2(1+x)\tag1$$
Now, we know that $\text{p}^2-1=(\text{p}+1)(\text{p}-1)$. So we can write:
$$(\text{p}+1)(\text{p}-1)=12x^2(1+x)\tag2$$
Suppose $x$ is an even number: $x=\text{m}\cdot2^\text{k}$ (where $\text{m}$ is odd). We get:
$$(\text{p}+1)(\text{p}-1)=12\left(\text{m}\cdot2^\text{k}\right)^2\left(1+\text{m}\cdot2^\text{k}\right)=3\cdot4^{1+\text{k}}\text{m}^2\left(1+\text{m}\cdot2^\text{k}\right)\tag3$$
Then we need to have $\text{m}^2\text{|}\text{p}\pm1$, such that $\text{p}=\pm1+\text{hm}^2$ (where $\text{h}\in\mathbb{Z}$). So we can write:
$$\text{h}^2\text{m}^2\pm2\text{h}=3\cdot4^{1+\text{k}}\text{m}^2\left(1+\text{m}\cdot2^\text{k}\right)\tag4$$
Which is the same as (because $2^x\cdot4^{1+x}=2^{2+3x}$ and $\text{m}\ne0$):
$$\text{h}^2\pm2\text{h}-3\cdot2^{2+3\text{k}}\text{m}-3\cdot4^{1+\text{k}}=0\tag5$$
Can someone help me finish the proof?
 A: Given a constant $\, D := 108,\,$ consider the function defined by
$$ f(x,y) := 9 + D\,x^2(1+x) - y^2 \tag{1} $$
where $\, f(x,y) = 0\,$ is the equation of an elliptic curve.
It is equivalent to the curve in Weierstrass form
$$ E: y^2 = x^3 + 1/D\,x^2 + 12/D^3 \tag{2} $$
in the following sense. $\,f(x,y) = 0 \,$ iff $\,[x/D,y/D^2]\,$ is
a point of curve $\,E\,$
which has $j$-invariant equal to $\,-12288/25.\,$
This curve is equivalent to the LMFDB 135.a1 curve
$$ E135a: y^2 + y = x^3 - 3x + 4. \tag{3} $$
The curve $\,E\,$ has a rational point $\,P:=[1/D,15/D^2]\,$ of rank $\,1\,$ which PARI/GP seems not to be able to provide, but it does
provide a rational generating point $\,[4,-8]\,$ for curve $\,E135a\,$.
Given a point $\,[x,y]\,$ of $\,E135a\,$ then
$\,[(x-1)/(3D),(-2y-1)/D^2]\,$ is a point of $\,E.\,$
Each point $\,[x,y]\,$ of $\,E\,$ satisfies 
$$ 0 = f(x\, D,y\, D^2). \tag{4} $$
However there are only a finite number of integer solutions to $\,f(x,y)=0.\,$
They correspond to the generator multiples
$$ 1P\mapsto(1,15),\, 2P\mapsto(0,3),\, 4P\mapsto(-1,-3),\, 7P\mapsto(6,-135),\,
 8P\mapsto(4,93). \tag{5}$$
Each solution $\,(x,y)\,$ yields a solution $\,(x,-y).\,$ There
are no other solutions in integers.
Some PARI/GP code is
D = 108; E = ellinit([0,1/D,0,0,1/12/D^2]); P = [1/D,15/D^2];
E135a = ellinit([0, 0, 1, -3, 4]);
print("E:",ellgenerators(E));
print("E135a:",ellgenerators(E135a));
for(n=1, 9, print(n," ",[x,y]=ellmul(E,P,n); [x*D,y*D^2]));

