Equality of ideals in $mathbb{Z}[sqrt{-7}]$

Let $J = (2, 1 + sqrt {-7})$. Obviously, $2 ∈ sqrt I$. So let’s investigate modulo $2$:

Let $S = R/2R$. For $x ∈ R$, since $2R ⊆ sqrt I $, $$x ∈ sqrt I iff [x]_{2R} ∈ sqrt I/2R.$$ However, $operatorname{char} S = 2$ – so $[(1 pm sqrt {-7}))^2]_{2R} = [8]_{2R} = 0$, so $1 pm sqrt{-7} ∈ sqrt I$. Therefore $J ⊆ sqrt I$.

But $R/J cong ℤ/2ℤ$, so $J$ is maximal, so $J = sqrt I$.