Note: it's not the maximal ideal but rather a maximal ideal, since for every prime $p$, $\langle p \rangle$ is maximal in $\mathbb Z$. To see this, note that $\langle p \rangle = p \mathbb Z \subsetneq \mathbb Z$ and if $\langle p \rangle \subsetneq I$ then there is an $i \in I$ such that $i$ is not a multiple of $p$. Since $p$ is prime you have $\gcd (p,i) = 1$ which means that there exists $k,n \in \mathbb Z$ such that $ki + np = 1 \in I$ and hence $I = \mathbb Z$.
Now to show that $I[x]$ is not maximal in $\mathbb Z[x]$ use Belgi's answer. You need to find a proper ideal $J$ of $\mathbb Z$ such that $I[x] \subsetneq J \subsetneq \mathbb Z[x]$.
Since $I = \langle 2 \rangle$, adding $\langle 2 \rangle $ to $I[x]$ yields $I[x]$ again, so your idea doesn't work in this case.
Maybe this helps: what is $I[x]$? It's all polynomials with even coefficients. To get an ideal $J$ that properly contains $I[x]$ you can adjoin another element. The element you adjoin cannot be in $\mathbb Z$ only since if it's even, you end up with $I[x]$ again and if it's odd you end up with $\mathbb Z$. So what you adjoin has to contain $x$.