Diophantine Equation Without Using Fermat's Last Theorem

I'm having trouble with a problem. The problem asks me to solve the equation $(x+1)^4-(x-1)^4=y^3$ in integers. I found out that the only integer solution is $(0,0)$. I found this answer by setting $x$ as $a^3$ and $x^2+1$ as $b^3$. After doing that, I got the equation $a^6+1^n=b^3$, which assures me that there is no other solution than $(0,0)$ by Fermat's Last Theorem. However, I just realized that I am not supposed to use Fermat's Last Theorem. So far, I have simplified the equation to $8x^3+8x=y^3$. Please help me prove that the only integer solution is $(0,0)$ without using Fermat's Last Theorem.

Answer

Consider $y^3=8x(x^2+1)$. Then $y$ must be even (say $y=2z)$. Then we get $z^3=x(x^2+1)$. Then we have
$$z^3-x^3=x.$$
OR

$$(z-x)(z^2+zx+x^2)=x$$
But the factor $|x^2+zx+z^2| \geq |x|$. So for this to be true $x=0=z$.