Prove that if $n \in \mathbb{Z}[\sqrt{2}]$ has an even norm, then $\sqrt{2} \mid n$ Aside from multiplying and dividing some specific numbers in this ring, e.g., $(1 + \sqrt{2})\sqrt{2}$ I have not really done anything productive on this question. I either go around in circles or jump to conclusions, such as that it follows from the fact that $\langle \sqrt{2} \rangle$ is a prime ideal with prime norm.
Though I do have this hunch that a very similar argument can be used to prove something similar in $\mathbb{Z}[\sqrt{-2}]$...

If the norm $N(a+b\sqrt 2)=a^2-2b^2$ is even, then $a^2$ is even, and so is $a$. Then $a=2a'$ for some integer $a'$. Therefore :

$$a+b\sqrt 2 = 2a'+b\sqrt 2=\sqrt 2 (b+a'\sqrt 2)$$ From this you conclude that $\sqrt 2 \mid a+b\sqrt 2$ in $\Bbb Z[\sqrt 2]$, provided that the norm of $a+b\sqrt 2$ is even.


More generally, if $d \in \Bbb Z$ is square-free, and if $d \mid N(a+b\sqrt d)$ (in $\Bbb Z$), then $\sqrt d \mid a+b\sqrt d$ in $\Bbb Z[\sqrt d]$.
Indeed, if $N(a+b\sqrt d)= a^2-db^2$ is a multiple of $d$, then $a^2$ is a multiple of $d$. As $d$ is square-free, you have that $d \mid a$ (this is not true in general, for instance $d=2 \cdot 4^2=32$ divides $a^2=64=8^2$, but $32$ doesn't divide $a=8$).
Then you can write $a=da'$ for some integer $a'$. Therefore $$a+b\sqrt d = \sqrt d (a'\sqrt d + b)$$ is a multiple of $\sqrt d$ as desired.

The converse also holds (if $d$ is square-free). Suppose that $x=a+b\sqrt d$ is a multiple of $\sqrt d$ (in $\Bbb Z[\sqrt d]$).
Then you can write $$a+b\sqrt d = \sqrt d(m+n\sqrt d)=nd+m\sqrt d$$
for some integers $m$ and $n$.
The norm of $x$ is
$$N(x)=N(nd+m\sqrt d)=n^2 d^2 - dm^2=d(dn^2-m^2).$$
In particular the norm is a multiple of $d$.

Here is an example. Let $d=6$ and $x=2+\sqrt 6$.
The norm of $x$ in $\Bbb Z[\sqrt 6]$ is $N(2+\sqrt 6)=4-6=-2$ which is not a multiple of $d=6$. In particular, $x$ is not a multiple of $\sqrt 6$ in $\Bbb Z[\sqrt 6]$.
It is possible to prove this without using norms.
Indeed, suppose that we could write
$$x=2+\sqrt 6 = \sqrt 6(m+n\sqrt 6) = 6n + m\sqrt 6 \in \Bbb Z[\sqrt 6]$$
for some integers $m$ and $n$.
Then recall that if $a+b\sqrt d = 0$ then $a=b=0$ (where $d$ is square-free). Suppose for a contradiction that $b≠0$. Then $\sqrt d =-a/b$ would be rational, which is impossible. Then $b=0$, and $a=0$.
In particular, $2+\sqrt 6 = 6n + m\sqrt 6$ is equivalent to $2-6n+\sqrt 6 (1-m)=0$. This yields $2-6n=0$ and $1-m=0$. But $3$ divides the integer $6n$ and $6n=2$... this is not possible.
Hence, we proved that $x=2+\sqrt 6$ is not a multiple of $\sqrt 6$ in $\Bbb Z [\sqrt 6]$.
(Notice that saying "$\sqrt d$ is irrational" is equivalent to saying $N(a+b\sqrt d)=0 \iff a=b=0$).
