Solving the diophantine equation $y^{2}=x^{3}-2$

The key is to factor this over $\mathbf{Z}[\sqrt{-2}]$ as $(y-\sqrt{-2})(y+\sqrt{-2})=x^3$. Working mod $8$ it is not too hard to see that $x$ and $y$ must be odd, and that $\gcd((y-\sqrt{-2}),(y+\sqrt{-2}))=1$; else every divisor would have an even norm, contradiction.

Therefore since $\mathbf{Z}[\sqrt{-2}]$ is a UFD and since the only units are $\pm 1$, each factor must be a cube. Thus we can write $$(y+\sqrt{-2})=(a+b\sqrt{-2})^3=(a^3-6ab^2)+(3a^2b-2b^3)\sqrt{-2}$$ Then, $y=a^3-6ab^2$ and $1=3a^2b-2b^3=b(3a^2-2b^2)$. From the two equations it is clear that $b=\pm 1\implies a=\pm 1$, so the only possible solutions are $(x,y)=(3,\pm 5)$.