I am having trouble solving the diophantine equation given in the title. This is how far I came:
We can factor in $\mathbb{Z}[i]$ $y^2+4=x^3\Rightarrow (y+2i)(y-2i)=x^3$. I want to show now that their gcd is a unit so in $\{\pm 1,\pm i\}$. Suppose $d=u+vi$ is the gcd, then $u+vi\text{ | }2y$ and $u+vi\text{ | }4i$. Now taking the norm, we find $u^2+v^2\text{ | }4y^2$ and $u^2+v^2\text{ | }16$. From the second part we find that $d=u+vi\in\{\pm 1,\pm i, \pm 4,\pm 4i,\pm 1\pm i,\pm 2\pm 2i\}$. Obviously $\pm 4$ and $\pm 4i$ cannot divide $y\pm 2i$. I don't know how to show it for the others.
If we assume that $y+2i$ and $y-2i$ are coprime and using the equation $(y+2i)(y-2i)=x^3$ we find that $y-2i=\epsilon \alpha^3$ and $y+2i=\delta \beta^3$ where $\epsilon$ and $\delta$ are units. From here on it is straightforward to find what $\alpha$ and $\beta$ can be. My only problem is showing that $y-2i$ and $y+2i$ are coprime.
Suppose $(a+bi)\mid (y-2i), (a+bi)\mid (y+2i) \to a+bi\mid 4i \to a^2+b^2\mid 16 \to a^2+b^2=1,2,4, 8, 16$. Can you take it from here?