Diophantine Equation $2x^2+25=y^3$

Question

I'm trying to find integer solutions to: $2x^2+25=y^3$.

Here's what I've managed to do so far:

1. y is odd. y and x are co-prime.
2. In $\mathbb{Q}(\sqrt{2},i)$ we can write: $(\sqrt{2}x+5i)(\sqrt{2}x-5i)=y^3$.
3. Using (1) I showed $(\sqrt{2}x+5i)$ and $(\sqrt{2}x-5i)$ are co-prime.
4. So assuming we have unique factorization (do we?) we have $a^3=\sqrt{2}x+5i$ and $b^3=\sqrt{2}x-5i$ for some $a,b$ in the ring of integers of $\mathbb{Q}(\sqrt{2},i)$.
5. I also know that an integral base in $\mathbb{Q}(\sqrt{2},i)$ is $\{1,\sqrt{2},i,\frac{1}{2}\sqrt{2}(1+i)\}$.
   I have tried to write $a$ in general form and get some restrictions on its coefficients but all I've reached is that $a\equiv -\sqrt{2}x+i(\text{mod } 3)$.

Any direction will be highly appreciated.