How do I prove $2$ is prime (or irreducible) in all $Z[\sqrt{d}]$ with $d < -7$? 
How do I prove $2$ is prime (or irreducible) in all $Z[\sqrt{d}]$ with $d < -7$?

I feel like the answer is right under my nose, but I just can't see it.

For convenience, let's say $-d$ instead of $d$, with $d \in \mathbb{Z}^+$ and $\mu(d) \neq 0$.
If 2 is composite or reducible in $\mathbb{Z}[\sqrt{-d}]$, that means $x^2 + dy^2 = 2$ has a solution in integers, or, if $d \equiv 3 \mod 4$, the equation $\frac{x^2}{4} + \frac{dy^2}{4} = 2$ holds. For the former, the only solutions are $1^2 + 1 \times 1^2$ and $0^2 + 2 \times 1^2$. And for the latter, $\frac{1^2}{4} + \frac{7 \times 1^2}{4}$. These solutions correspond to the factorizations $(1 - i)(1 + i)$, $(-1)(\sqrt{-2})^2$ and $\left(\frac{1}{2} - \frac{\sqrt{-7}}{2}\right)\left(\frac{1}{2} + \frac{\sqrt{-7}}{2}\right)$.
If $d > 9$ and $y \neq 0$, then clearly $x^2 + dy^2 \geq 10 > 2$ or $\frac{x^2}{4} + \frac{dy^2}{4} \geq \frac{5}{2} > 2$.
