Answer: $(x,y)=(-1,0)$ Notice that $$(x+2)^4-x^4=8(x+1)((x+1)^2+1)$$ Notice that both factors are relatively prime so $(x+1)^2+1$ is a perfect cube, and we are done by Catalan.

Write $t=x+1$, so the equation is $(t+1)^4 - (t-1)^4 = y^3$. It is equivalent to $8t^3 + 8t = y^3$, so necessarily $t(t^2+1)$ is a cube; but since both factors are coprime, they must both be cubes - hence with $t=a^3$ we must have $a^6 + 1 = b^3$ for some $a$ and $b$, so $(b-a^2)(b^2 + a^2b + a^4) = 1$ and solving $b - a^2 = b^2 + a^2b + a^4 = \pm 1$ yields $a=0$ and $b=1$ as the only solution - correspondingly $x=-1$ and $y=0$.