I am stuck on the following question which is given as follows:
Prove that the only integer solutions to the equation \begin{equation} x^2 + 13 = y^3 \end{equation} are $(70,17)$ and $(-70, 17)$.
(Hint: first show that $x$ is even and $y$ is odd)
I have seen a solution to this and first part to it is as follows.
Let $x$ and $y$ satisfy the given equation. A odd square is $1$ modulo $4$ and a cube cannot be $2$ modulo $4$ so $x$ is even, which immediately implies $y$ is odd.
So we have a factorisation of ideals $(x + \sqrt{-13}) \cdot (x - \sqrt{-13}) = (y)^3. \ $ The gcd of $(x + \sqrt{-13})$ and $(x - \sqrt{-13})$ divides $(2\sqrt{-13}) = (2, 1+\sqrt{-13})^2 \, (\sqrt{-13}). \ $ As $y$ is odd, $(y)$ cannot be divisible by $(2, 1+\sqrt{-13})$ hence \begin{equation} (x + \sqrt{-13}, \, x - \sqrt{-13}) = (1) \hspace{5pt} \text{or} \hspace{5pt} (\sqrt{-13}) \hspace{5pt} (*) \end{equation} In the latter case, $x$ would be divisible by $13$, hence $y$ would be divisible by $13$ but then $13 \ \big| \ y^3$ and $x^2 + 13 \equiv 13 \pmod{13^2}$ which is a contradiction so $(x + \sqrt{-13}, \, x - \sqrt{-13}) = (1)$. It follows that the principal ideal $(x + \sqrt{-13})$ is the cube of an ideal $I$.
So firstly why does $(*)$ hold given that $(y)$ cannot be divisible by $(2, 1 +\sqrt{-13})$ and in the latter case why does this mean $x$ is divisible by $13$?. Finally how do we know that $(x + \sqrt{-13})$ is the cube of an ideal $I$?
Any help would be very much appreciated!