I just want to make sure I am getting this correct, as I am just learning ring theory and quadratic integer rings.
Let $\alpha = 7 + \sqrt{-14}$.
To show that $\alpha \notin (7)$ I will show a contradiction. If we assume that $\alpha \in (7)$, then there must exist some element $q \in \mathbb{Z}[\sqrt{-14}]$ such that $7q=\alpha$. Let $q = c + d\sqrt{-14}$, then this amounts to finding $c$ and $d$ $\in \mathbb{Z}$ such that $$(7+0\sqrt{-14})(c+d\sqrt{-14})=(7+\sqrt{-14})$$ . Since $7c +0d\sqrt{-14}=7$, this implies that $c = 1$. We also must have that $7d\sqrt{-14} + 0c\sqrt{-14} = 1\sqrt{-14} \implies 7d=1$ which is impossible for $d \in \mathbb{Z}$.
So we must not assume that $\alpha \in (7)$, so $\alpha \notin (7)$.