Statement of Problem : "Find all integers $n \in \mathbb{Z}$ so that $\sqrt{n} \in \mathbb{Q}[\sqrt{2}]$.
I think that it is just the integers that can be written as the square of another integer or $\{ n \vert n = a^2 \space \space \forall a \in \mathbb{Z}\}$, as elements of that fields extension are of form $\{a + b\sqrt{2} \vert a,b \in \mathbb{Q}\}$, so $ n = a^2 + 2\sqrt{2}ab + 2b^2$, and so either a or b is 0, and the case where $a = 0$ is a proper subset of the case where $b = 0$, so it's just all the integers that are of form $n = a^2$ such that $a = \frac{p}{q}$ such that $q=1$. Is this correct?
$$ n=a^2+2\sqrt{2}ab+2b^2 \Rightarrow a=0 \ \ \text{or} \ \ b=0 .$$
Therefore $$ n\in\{ 2b^2:b\in\mathbb{Z} \}\cup\{a^2:a\in\mathbb{Z}\} .$$