How can I solve $x^2+2=y^3$ in $\mathbb{Z}$?

Question

Prove that $\left \{ (x,y)\in\mathbb{Z}^2:x^2+2=y^3 \right \}\subseteq \left \{ (-5,3),(5,3) \right \}$.

Answer 1

First note that $x$ must be odd.

Then show that in $R=\mathbb Z[\sqrt {-2}]$ that $x+\sqrt{-2}$ and $x-\sqrt{-2}$ are relatively prime. Since $R$ is a unique factorization domain, and all units are cubes, this means the $x+\sqrt{-2}$ is a cube in $R$ (Since their product is a cube.)

Then solve:

$$x+\sqrt{-2}=(a+b\sqrt{-2})^3.$$