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.

One might argue in this fashion, instead: Suppose that $J$ is an ideal such that $I\subsetneq J\subseteq\Bbb Z[x],$ and take any $y\in J\setminus I.$ By definition of $I,$ we know that $x$ doesn't divide $y,$ and so if $y=px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ then $n\ne 0.$ In fact, $n$ cannot be any even integer. (Why?) Thus, if $y=px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ then there is some $k\in\Bbb Z$ such that $y=px+2k+1.$ Finally, we note that $px+2k\in I,$ whence $px+2k\in J,$ and so $1=y-(px+2k)\in J,$ and so $J=\Bbb Z[x].$
All that remains is to justify the steps above, and to demonstrate that every element of $\Bbb Zx$ can be written in the form $px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ which I leave to you.
