$y^2=x^3-2$

We want to solve the equation $y^2=x^3-2$ over the integers.

By writing $y^2+2=x^3$ we can factor on $\mathbb{Z}[\sqrt{-2}]$ as

$$(y-i\sqrt{2})(y+i\sqrt{2})=x^3.$$

Using congruences modulo $8$, one can show that both $x,y$ must be odd, and it can also be shown that $(y-i\sqrt{2})$ and $(y+i\sqrt{2})$ are relatively prime (if it were not the case, any divisor would have even norm, which is not possible).

Therefore, by unique factorization, and using that the only units (http://planetmath.org/UnitsOfQuadraticFields) on $\mathbb{Z}[\sqrt{-2}]$ are $1,-1$, we have that each factor must be a cube.

So let us write

$$(y+i\sqrt{2})={(a+bi\sqrt{2})}^3=(a^3-6a{b}^2)+i(3a^{2}b-2b^3)\sqrt{2}$$

Then $y=a^3-6a{b}^2$ and $1=3a^{2}b-2b^3=b(3a^2-2b^2)$. These two equations imply $b=\pm 1$ and thus $a=\pm 1$, from where the only possible solutions are $x=3,y=\pm 5$.