Factoring 2 in $\mathbb{Q}(\sqrt{-7})$

$$2=\frac{1+\sqrt{-7}}{2}\frac{1-\sqrt{-7}}{2}$$ $$3=\frac{1+\sqrt{-11}}{2}\frac{1-\sqrt{-11}}{2}$$ $$5=\frac{1+\sqrt{-19}}{2}\frac{1-\sqrt{-19}}{2}$$ $$11=\frac{1+\sqrt{-43}}{2}\frac{1-\sqrt{-43}}{2}$$

See the pattern ?

Since 2 is not pure imaginary you must, for non zero $b,d$, have $a=c$ and $d=-b$. This reduces to solving $(\frac{a+b\sqrt{-7}}{2})(\frac{a-b\sqrt{-7}}{2})=2$, so you need to find $a,b$ such that $a^2+7b^2=8$.

Your mileage may vary...

I find these problems much easier if instead of trying to solve $(a - b \theta)(a + b \theta)$ (where $\theta = \frac{1 + \sqrt d}{2}) = p$, I try to solve $$\left( \frac{a - b \sqrt d}{2} \right) \left( \frac{a + b \sqrt d}{2} \right) = p.$$ Then $$\left( \frac{a - b \sqrt d}{2} \right) \left( \frac{a + b \sqrt d}{2} \right) = \frac{a^2 + (-d)b^2}{4}$$ and $a^2 + (-d)b^2 = 4p$.

So then, for $d = -7$, $p = 2$, I solve $a^2 + 7b^2 = 8$. The answer then becomes obvious: $a = 1$, $b = 1$ also.

With the thetas, the whole thing is kind of confusing.

$$2 = \frac{1 + 7}{4} = \left( \frac{1 + \sqrt{-7}}{2} \right) \left( \frac{1 - \sqrt{-7}}{2} \right) = (1 - \theta) \theta,$$

(adjust $\theta$ as you go on to each of these)

$$3 = \frac{1 + 11}{4} = \left( \frac{1 + \sqrt{-11}}{2} \right) \left( \frac{1 - \sqrt{-11}}{2} \right) = (1 - \theta) \theta,$$ $$5 = \frac{1 + 19}{4} = \left( \frac{1 + \sqrt{-19}}{2} \right) \left( \frac{1 - \sqrt{-19}}{2} \right) = (1 - \theta) \theta,$$ $$11 = \frac{1 + 43}{4} = \left( \frac{1 + \sqrt{-43}}{2} \right) \left( \frac{1 - \sqrt{-43}}{2} \right) = (1 - \theta) \theta,$$

which is to say, for your $(a + b \theta)(c + d \theta)$, we have $a = 1$, $b = -1$, $c = 0$, $d = 1$. The thetas obscure your approach to $4p$. But, like I said, your mileage might vary.