The statement goes as the title. So I figured that $29 = (5+2i)(5-2i)$ in $\mathbb{Z}[i]$. However I'm not sure how to continue since the respective rings look like this:
$\mathbb{Z}[i\sqrt{ 2}]=\{a+bi\sqrt{2}:a,b\in \mathbb{Z}\}$ and $\mathbb{Z}[\sqrt{7}]=\{a+b\sqrt{7}:a,b\in \mathbb{Z}\}$.
You correctly used the fact that for a factorization into 2 factors in $\mathbb{Z}[i]$ you need a representation of 29 as the sum of 2 squares. So for such a factorization in $\mathbb{Z}[i\sqrt 2]$ you need a representation of 29 as the sum $a^2+2b^2$. And for such a factorization in $\mathbb{Z}[\sqrt 7]$ you need a representation of 29 as the sum $a^2-7b^2$.