Let $K=\mathbb{Q}(i,\sqrt{2m})$ where $m \in \mathbb{Z}$ is odd and squarefree. Let $\alpha = (1+i)\sqrt{2m}/2$. Then $\alpha^2=im$, such that $\alpha$ is part of the ring of integers $\mathcal{O}_K$.
We need to show that for $\mathfrak{B}=(1+i,1+\alpha)$ we have
$$2\mathcal{O}_K=\mathfrak{B}^4$$
I tried to calculate $\mathfrak{B}^2$ first, but my computations didn't work.
I am aware that one can take linear combinations of the different generators by elements generated by the integral basis $\{1, \alpha,i, \sqrt{2m}\}$. I also tried to introduce the substitution $\alpha^2=im$ at multiple places. Yet, I keep getting stuck in the calculations. Maybe I am missing some elegant trick?
When somebody is totally ramified, Eisenstein is your friend :)
The minimal polynomial of $1+\alpha$ (over $\mathbb Q$) is $(x-1)^4+m^2$ and it is $2$-Eisenstein. As a result, the prime ideal $\mathfrak P=(2,1+\alpha)$ satisfies $\mathfrak P^4=(2)$. Since $2i=(1+i)^2$ and $\mathfrak P$ is prime, we have also $(1+i)\in\mathfrak P$, i.e. $\mathfrak P=\mathfrak B$.