Let $I = \{a+ ib \in \Bbb Z[i] : 2 \mid a-b\}$ then $I$ is a maximal ideal of $ \Bbb Z[i]$.
We consider an ideal $J$ such that $I \subset J\subset\Bbb Z[i] $. So there exists an element $p \in J$ but not in $I$. Note that $a \not\mid a-b$ when $a,b$ are of odd parity i.e. one is odd and other is even. Without loss of generality we assume that $p = c+ id$ where $c$ is odd and $d$ is even. Then we have $q = (c-1)+id \in I$ since both $c-1,d$ are even and hence $q = (c-1)+id \in J$. Now since $J$ is an ideal, so $p-q = 1 \in J$. Thus $J =\Bbb Z[i] $.
Hence $I$ is a maximal ideal of $ \Bbb Z[i]$.
Is the proof correct??
yes your proof is absolutely correct