Factoring 2 in $\mathbb{Q}(\sqrt{-7})$ $7$ is a Heegner number, so all numbers in $\mathbb{Q}(\sqrt{-7})$ have a unique factorization.  
I'm told that:   


*

*$2$ is not prime in $\mathbb{Q}(\sqrt{-7})$,

*$3$ is not prime in $\mathbb{Q}(\sqrt{-11})$,  

*$5$ is not prime in $\mathbb{Q}(\sqrt{-19})$,  

*$11$ is not prime in $\mathbb{Q}(\sqrt{-43})$,  

*$17$ is not prime in $\mathbb{Q}(\sqrt{-67})$,  

*$41$ is not prime in $\mathbb{Q}(\sqrt{-163})$.   


For the first of these, I've tried finding prime numbers $(a+b (1+\sqrt{-7})/2)$ and $(c+d (1+\sqrt{-7})/2)$, with product $2$, but I haven't had luck so far.

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.
