Find all integer pairs $(x, y)$ for which

$$
(x+2)^{4}-x^{4}=y^{3}
$$
$$
(x+2)^{4}-x^{4}=y^{3} \text {. }
$$

Introducing the notations $x+1=u$ and $y / 2=v$, our equation transforms into:

$$
u^{3}+u=v^{3}
$$

If equation (1) holds for integers $x, y$, then from (2) it is clear that $v^{3}$ is an integer, and $v$ is rational (since it is half of an integer), which is only possible if $v$ itself is an integer. Therefore, we will only look for integer solutions to (2). 

$$
(u+1)^{3}-\left(u^{3}+u\right)=\frac{(3 u+1)^{2}}{3}+\frac{2}{3}>0
$$

and

$$
\left(u^{3}+u\right)-(u-1)^{3}=\frac{(3 u-1)^{2}}{3}+\frac{2}{3}>0
$$

due to these identities, for every $u$,

$$
(u+1)^{3}>u^{3}+u>(u-1)^{3}
$$

thus $u^{3}+u$ can only be equal to a cube number if $u^{3}+u=u^{3}$, i.e., $u=v=0$. In this case, $x=-1$, and $y=0$, which is the only integer solution to (1).