To prove that $I = \{\,(n,m) \in \Bbb Z \times \Bbb Z : n,m $ are even $\}$ is not a maximal ideal of $ \Bbb Z \times \Bbb Z $. To prove that $I = \{\,(n,m) \in \Bbb Z \times \Bbb Z : n,m $ are even $\}$ is not a maximal ideal of $ \Bbb Z \times \Bbb Z $.
Thus there exists an ideal $J$ of $ \Bbb Z \times \Bbb Z $ such that $I \subset J \subset  \Bbb Z \times \Bbb Z $. 
Will $J = \{\,(n,m) \in \Bbb Z \times \Bbb Z : n $ is even $\}$ work here?

Yes, it will. It is clearly between $I$ and the full ring and is an ideal.
Alternatively, without exhibiting $J$, you can calculate $(1,0)\cdot(0,1)=(0,0)$, which  shows that the product of two elements $\notin I$ can be $\in I$; hence $I$ is not even prime, let alone maximal.
