I'm not sure how to answer this question. If $p(x)=(x-a)(x-b)$ where $a,b \in \mathbb{Z}$ then $(p(x)) \subset (x,a)$ and $(p(x)) \subset (x,b)$ but how do I show that $(x,a)$ and $(x,b)$ are maximal?
Is the original claim true? Hints will be appreciated.
By Zorn's Lemma there exists a maximal ideal $I$ in $\mathbb{Z}[x]$ such that $x-a\in I$. For the same reason, there exists $J$ maximal such that $x-b\in J$. Of course $(p)\subset I$ and $(p)\subset J$.
Now I should prove that any other maximal ideal $K$ such that $(p)\subset K$ then $K=I$ or $K=J$. In deed, if $(p)\subset K$ then $(x-a)(x-b)\in K$. As $K$ is prime (every maximal ideal is a prime ideal in an unital commutative ring), then $x-a \in K$ or $x-b \in K$. In the first case we should have $K=I$ since $I$ is the maximal ideal such that $x-a$ is in it. In the second case, $K=J$.
Therefore there are only two maximal ideals such that they contain $(p)$.