I want to show that the equation $x^3=y^2+68$ has no integer solutions by considering ideals in the ring of integers of $K=\mathbb{Q}(\sqrt{-17})$ and applying results like the Dedekind-Kummer Theorem.
I start with converting it to an equation of ideals in $\mathcal{O}_K$: $\langle x \rangle^3=\langle y+2\sqrt{-17}\rangle\langle y-2\sqrt{-17}\rangle$.
Then I try and show that the two ideals on the right hand side are coprime, in order to deduce that each is a cube. This is where I get stuck.
Normally, the following method works: Suppose for a contradiction that some prime ideal $\mathfrak{p}$ divided both of the ideals $\langle y+2\sqrt{-17}\rangle$ and $\langle y-2\sqrt{-17}\rangle$. Then $4\sqrt{-17}=(y+2\sqrt{-17})-(y-2\sqrt{-17})\in\mathfrak{p}$, so $\mathfrak{p}$ divides $\langle 4\sqrt{-17}\rangle$. Comparing ideal norms then gives Nm$(\mathfrak{p}) | 16\cdot 17$. The norm of a prime ideal is always some prime power, so either Nm$(\mathfrak{p})$ is even, or 17.
17 gives a contradiction, as Nm$(\mathfrak{p})$ divides Nm$(\langle y+2\sqrt{-17}\rangle) = y^2+68$, so this would imply that $17|y^2$, $17|x^3$ and thus $17|y$, $17|x$, so $17^2|x^3=y^2+68$ and then $17^2|68$, a contradiction.
But I am struggling to reach a contradiction in the case where Nm$(\mathfrak{p})$ is even. Any tips would be appreciated! Once I am able to show that the two ideals are coprime then I am able to finish the argument, so it is just this step that I'm struggling with.