Show that the ideal $ (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal Show that the ideal $ I =  (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal.

My thoughts so far:

Work by contradiction. Assume that $ I $ is principal, i.e. that it is generated by some element $ z = a + b\sqrt{-7} \in \mathbb{Z}[\sqrt{-7}] $. I'm really not sure what to consider though - I can't really 'see' what $ I $ looks like.

Any help would be greatly appreciated. Thanks
Put $\rm\, w = 1\!+\!\sqrt{-7}.\,$ By norms $2\,$ is irreducible so, if principal $\rm\,\color{#c00}{ (2,w) = (1)}\ $ [not $(2)$ by $\rm\,2\nmid w$]  

so $\rm\ 2\mid 2w,\,ww'\Rightarrow\ 2\mid (2w',ww') = \color{#c00}{(2,w)}(w') =  (w'),\ $ so $\rm\ 2\mid w',\ $ contradiction. $\ \, $ QED

This is a special case of the fact that the failure of an irreducible element to be prime (or a failure of Euclid's Lemma) immediately yields a nonexistent gcd and nonprincipal ideal - see this answer.
To prove this note that $1 \not \in I$ so $I \not = (1)$. Then suppose $I = (\alpha)$, that implies that $\alpha = 2$ or $\alpha = 1 + \sqrt{-7}$ since those are both irreducibles, but neither of them can hold since one irreducible is not a multiple of another.