number theory - Given integers $x,,y$ s.t. $x^2-16=y^3$, show that $x+4$ and $x-4$ are perfect cubes

Suppose $x$ and $y$ are some integers satisfying $$x^2-16=y^3.$$ I'm trying to show that $x+4$ and $x-4$ are both perfect cubes.

I know that the greatest common divisor of $x+4$ and $x-4$ must divide $8$, but I don't know where to go from there. Would anyone be able to help?