Prove by definition that $(x,2)\subset\mathbb Z[x]$ is a maximal ideal When  the  polynomial  ring  $\mathbb{Z}[x]$  is  quotiented  by  the  ideal  $(2,x)$  we  get  a   field  as $\mathbb{Z}[x]/(x,2)\cong\mathbb{Z}/(2)\cong\mathbb{Z}_{2}$ which  is  a  field. 
But  I  want  to  prove  it  with  the  definition  of  maximal  ideal. So let  $I$=$(2,x)$  be  not  maximal. If  possible, there  is ideal $J$  such  that $I\subsetneq J\subsetneq \mathbb{Z}[x]$ . If  only  $\mathbb{Z}[x]$  were  a  PID,  I  could  say $J$  should  have  one  element  dividing  both $2$  and $x$  but  their  gcd  being  $1$ , $1\in J$  i.e. $J$=$\mathbb{Z}[x]$. But  since  $\mathbb{Z}[x]$  is  not  a  PID   this  argument  won't  work  ,  $J$  can be  generated  by  two  different  elements .  So  how  do  I  contradict  the  existence  of  $J$  here?
Thanks  for  any  help.

Try it like this.  Let $p(x)= a_0 + a_1x + a_2x^2 + \dots + a_nx^n$ (with each $a_k \in \mathbb{Z}$) be any element that is in $J$ but not in $I$.  Now note that $a_1x + a_2x^2 + \dots + a_nx^n$ is already in $I$ (because it is a multiple of $x$), hence it must be true that $a_0 \in J$ but $a_0 \notin I$.  Now $a_0$ can't be an even integer -- if it were, it would be in $I$ -- so it must be an odd integer.  From this, it is easy to conclude that $1 \in J$, and you are done. 
