 Solution

A Euclidean domain is an integral domain $R$ equipped with a function $\delta :R\setminus \{0\}\to \mathbb{N}$ such that for $a,b\in R$, $b\ne 0$, there are $q,r\in R$ withFor the Gaussian integers $\mathbb{Z}[i]$, take $\delta (z)=N(z)=|z{|}^2$. Choosing a Gaussian integer nearest to $a/b$ makes the remainder norm smaller than $N(b)$.

The units are precisely the elements of norm one:Unique factorization in this Euclidean domain givesThe displayed factors have prime norms $2$ or $5$, so they are irreducible; factors appearing together are nonassociate.

Now suppose $x^2+4=y^3$. Necessarily $y>0$. First let $y$ be odd. Then $x$ is odd, and $x+2i$ and $x-2i$ are coprime in $\mathbb{Z}[i]$: a common Gaussian prime would divide $4i$, while their product has odd norm. Henceup to a unit, which can be absorbed into the cube. Comparing imaginary parts givesChecking $b\mid 2$ yields only $b=-2$, $a=\pm 1$, and therefore

If $y$ is even, congruence modulo $8$ gives $x=2X$ with $X$ odd and $y=2Y$, whereEach of $X\pm i$ contains exactly one factor $1+i$, so the coprime quotients are cubes up to units. ThusComparing the imaginary part after the four possible units reduces toEach integer factor must have absolute value one. Substitution then gives $a^2+b^2=1$, so $Y=1$ and $X=\pm 1$. HenceAll four pairs satisfy the equation, so the complete answer is