If $g(X)\in \Bbb Z[X]$ irreducible polynomial, then $\langle g(X) \rangle \trianglelefteq \Bbb Z[X]$ is not maximal I would like to prove that given an irreducible polynomial $g(X)\in \Bbb Z[X]$, then the ideal $\langle g(X) \rangle \trianglelefteq \Bbb Z[X]$ is not maximal. 

One can think to prove
$$\langle g(X) \rangle \subsetneq \langle p,g(X)\rangle \subsetneq \Bbb Z[X]$$
and maybe to use the ring isomorphism 
$$\frac{\Bbb Z[X]}{\langle p,g(X) \rangle}\cong\frac{\Bbb Z_p[X]}{\langle \overline g(X) \rangle }$$
where $p$ is a prime number and $\overline g(X)$ is $g(X)$ with coefficients in $\Bbb Z_p$. But how could we proceed? 

I face difficulty to prove $\langle p,g(X)\rangle \subsetneq \Bbb Z[X]$.

I know that this may be an easy question, but I have stuck. Also, any other ideas are welcome!
Hint: $\langle g\rangle$ cannot contain any elements with degree lower than the degree of $g$ (apart from the zero polynomial).
Taking into consideraction Arthur's and Wuestenfux's answers, I ll try to write down the full answer.

At first, if $\deg g(X)=0\implies g(X):=m\in \Bbb Z$, then $\Bbb Z[X]/\langle m \rangle \cong \Bbb Z_m[X]$, which is not a field, so $\langle g(X)\rangle $ is not maximal. 

Now, we will assume that $\deg g(X)\geq 1$. 


$\langle g(X)\rangle \subsetneq \langle p ,g(X) \rangle$, because if so, $p\in \langle g(X)\rangle \implies p=q(X)g(X),\ q(X) \in \Bbb Z[X \implies \deg (p)=\deg q(X)+\deg g(X) \implies 0=d_1+d_2,$ where $d_1,d_2\in \Bbb N$ and $d_2\geq 1$, contradiction. 
$\langle p ,g(X) \rangle \subsetneq \Bbb Z[X]$: We will show that 
$$1\notin \langle p ,g(X) \rangle  \iff \langle p ,g(X) \rangle  \neq \Bbb Z[X].$$
We suppose, contrary, that  $1\in \langle p ,g(X) \rangle$. Then,
\begin{alignat*}{2}
1\in \langle p ,g(X) \rangle \iff 1=a(X)p+b(X)g(X), \tag{*}
\end{alignat*}
for some $a(X),b(X)\in \Bbb Z[X]$. We consider the following cases:


(1) If $a(X),b(X)=0$, we have immediately contradiction, through (*).

(2) If $a(X)\neq 0,\ b(X)=0$, then $1=b(X)g(X) \implies \deg g(X)=0$, contradiction.

(3) If $b(X)\neq 0,\ a(X)=0$, then $1=a(X)p \implies p|1$, contradiction.

(4) If $a(X)\neq 0,\ b(X)\neq 0$, then if we take degrees in (*), we will have
$$0=\max\{ \deg(a(X)p),\underbrace{\deg (b(X)g(X))}_{\geq 1} \},$$
so again we have a contradiction.