$\langle x,2 \rangle$ is a maximal ideal in $\mathbb{Z}[x]$ Let $\mathbb{Z}[x]/\langle x,2 \rangle$. We know $\langle x,2 \rangle = \{f \in \mathbb{Z}[x] : f(0)$ is even integer$\}$
Also $$\mathbb{Z}[x]/\langle x,2 \rangle = \{f(x) + \langle x,2 \rangle\} = \{a + \langle x,2 \rangle\}$$ since we can write $$f(x) = a_0 + \cdots+ a_nx^n = a_0 + x(a_1 + \cdots + a_nx^{n-1}) = a + 2k + x(a_1 +\cdots + a_mx^{n-1}) = a + 2k + xg(x)$$ and since $\langle x,2 \rangle$ is an ideal, it absorbs $2k + xg(x)$. So, my approach would be to show that $\mathbb{Z}[x]/\langle x,2 \rangle$ is an integral domain. This would give that $\langle x,2 \rangle$ is prime and maximal. Am I correct so far?
 A: MSE--sometimes I just can't stop typing!
My way of looking at this is very similar to Citizen's, maybe redundantly so; like I said, sometimes I can't stop typing, so please bear with me . . . 
Let us denote the ideal $\langle 2, x \rangle$ of $\mathbf Z [x]$ by $J$; thus $J =  \langle 2, x \rangle \subset \mathbf Z [x]$.
First off, notice that the ideal $J = \langle 2, x \rangle$ of $\mathbf Z [x]$ consists of all polynomials of the form $2g(x) + xh(x)$ with  $g(x), h(x) \in \mathbf Z [x]$.  Next, observe that we can always express a polynomial of the form $2g(x) + xh(x)$ as $xp(x) + 2m$ for some $p(x) \in \mathbf Z [x]$ and $m \in \mathbf Z$.  Then two polynomials $q_1(x), q_2(x) \in \mathbf Z [x]$ represent the same element of $\mathbf Z [x]/J$ precisely when $q_1(x) - q_2(x) \in J$, that is when $q_1(x) - q_2(x) = xp(x) + 2m$ with $p(x)$, $m$ as above.  But since $xp(x)$ is an arbitrary polynomial with constant term $0$, it follows that $q_1(x) + J = q_2(x) + J$ exactly when the constant terms of $q_1(x)$, $q_2(x)$, which are $q_1(0)$, $q_2(0)$ respectively, differ by an even integer, i.e. $q_1(0) - q_2(0)$ is a mutiple of $2$.  Thus $q_1(0)$ and $q_2(0)$ are either both even, or both odd.  From this it follows that there are precisely two equivalence classes in $\mathbf Z [x]/J$, one containing all polynomials $q(x) \in \mathbf Z [x]$ with $q(0)$ even, which is in fact $J$ itself, and the other consisting of those  $q(x) \in \mathbf Z [x]$ with $q(0)$ odd, which is $1 + J$.  The natural projection homomorphism $\pi: \mathbf Z [x] \to \mathbf Z [x]/J$ thus effectively maps $\mathbf Z [x]$ onto a ring with precisely two elements, and since the product $q_1(x)q_2(x)$ of two polynomials $q_1(x), q_2(x) \in \mathbf Z [x]$ with $q_1(0), q_2(0)$ odd again has $q_1(0)q_2(0)$ odd, we have $(1 + J)^2 = (1 + J) \ne 0$ in $Z [x]/J$, which shows that multiplication in the quotient ring $Z [x]/J$ is precisely the same (up to isomorphism) as that in the two-element field $\Bbb F_2$.  Thus we see that $Z [x]/J$ is in fact a field itself, whence $J = \langle 2, x \rangle$ is a maximal ideal in $\mathbf Z [x]$.
Now in fact the maximality of $J$ can actually be seen directly once we have established that the ideal $\langle 2, x \rangle$ consists of precisely those polynomials of the form $xp(x) + 2m$, for any ideal $J'$ with $J \subset J'$, $J' \ne J$, must contain some $q(x) \in \mathbf Z [x]$, $q(x) \notin J$.  But we have shown that any such $q(x)$ must have $q(0)$ odd; but then we would have $1 \in J'$, whence
$J' = \mathbf Z [x]$, showing $J$ is maximal in $\mathbf Z [x]$.
It goes on (still can't stop typing!):  apparently for any $n \in \mathbf Z$ we have $\mathbf Z [x]/\langle n, x \rangle \simeq \mathbf Z_n$; this follows since, as in the above, $J = \langle n, x \rangle$ consists of those polynomials of the form $ng(x) + xh(x) = xp(x) + mn$; now  $q_1(x) - q_2(x) \in J$ means $q_1(0) - q_2(0) = mn$ for some $m \in \mathbf Z$; thus $q_1(0) \equiv q_2(0)\mod n$; as in the case $n = 2$, 
$q_1(x) + J = q_2(x) + J$ if and only if $q_1(0), q_2(0)$ are in the same residue class modulo $n$; furthermore the natural projection $\pi:\mathbf Z[x] \to \mathbf Z [x] / J$ is easily seen to map the polynomial $q_1(x)q_2(x)$ to $q_1(0)q_2(0) + J$.  Thus there are precisely $n$ elements in $\mathbf Z [x] / J \simeq \mathbf Z_n$.  
For primes $p$, we can also argue the maximality of $J = \langle p, x \rangle$ via
the observation that the elements of $J$ are of the form $xq(x) + mp$; then if $r(x) \notin J$ it must be of the form $r(x) = xs(x) + t$ where $(t, p) = 1$; but then any ideal $K$ containing $J$ and $r(x)$ must contain $1$, since there exist integers $a, b$ with $at + bp = 1$; as such $K = \mathbf Z [x]$ and thus $J$ is maximal.
If, in the above, $n$ is taken to be a prime, the we have $\mathbf Z [x] / J \simeq \mathbf Z_n = \Bbb F_n$, a field, and the ideal $J = \langle x, n \rangle$ is maximal in $\mathbf Z [x]$.  When $n$ is composite, however, $\mathbf Z_n$ is not a field and
$J$ will not be maximal.  If $n = rs$ with $r, s > 1$, then we have $\langle n, x \rangle \subset \langle r, x \rangle \subset \mathbf Z [x]$, with no equalities between these sets.  
OK, fingers tired; brain dead; can stop typing now.
I really hope this one helps someone, somewhere.
I say, "Cheers"; exhausted, I proclaim
Fiat Lux!
A: On the urging of @Julian Kuelshammer, I'm making an answer out of my comments. Since maximal ideals are prime it suffices to prove that $(x,2)$ is maximal, and for this it is enough to prove that $R:=\mathbf{Z}[x]/(x,2)$ is a field. What field should it be? Well, the image of $2$ in this ring is $0$ since $2\in(x,2)$, so if this is a field, it has to be of characteristic $2$. In fact the canonical map $\mathbf{Z}\rightarrow R$ has kernel containing $2\mathbf{Z}$, and in fact this must be the kernel, since $2\mathbf{Z}$ is a maximal ideal of $\mathbf{Z}$, so we get an induced ring map $\mathbf{F}_2=\mathbf{Z}/2\mathbf{Z}\rightarrow R$. It is injective since $2\mathbf{Z}=\ker(\mathbf{Z}\rightarrow R)$, or if you prefer, because $\mathbf{F}_2$ is a field. We claim that it is also surjective. Well, given $f(X)=a_nX^n+\cdots+a_1X+a_0\in\mathbf{Z}[X]$, we have $f(X)\equiv a_0\pmod{X}$. But the image of $a_0+2\mathbf{Z}\in \mathbf{F}_2$ under our map $\mathbf{F}_2\rightarrow R$ is precisely $a_0+(x,2)=f(X)+(x,2)$. So indeed the map is surjective, and therefore an isomorphism. 
