Since $\Bbb Z[\sqrt{3}]\cong \Bbb Z[x]/(x^2-3)$, you are looking for the maximal ideals of $\Bbb Z[x]$ which contain $(x^2-3)$. The maximal ideals of $\Bbb Z[x]$ appear among the prime ideals described here, and are exactly the ones generated by two elements.
There are three possibilities that each prime $p$ can fall under:
$x^2-3\pmod{p}$ is irreducible, as in the case when $p=5$. In that case, $(p,x^2-3)$ is maximal. This corresponds to the case when $3$ is not a quadratic residue in $\Bbb F_p$.
$x^2-3\equiv (x-\alpha)(x-\beta) \pmod{p}$, where $\alpha\neq\beta$, as in the case for $p=11$. We have two distinct maximal ideals $(p,(x-\alpha))$ and $(p,(x-\beta))$. This corresponds to the case when $3$ is a quadratic residue in $\Bbb F_p$.
$x^2-3\equiv (x-\alpha)^2 \pmod{p}$, as in the case $p=2$ and $p=3$. (Actually, you should show that these are the only two primes for this case.) Then $(p,(x-\alpha))$ is maximal.
For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M \subset F[x] \tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) \tag 2$
for some
$m(x) \in F[x]; \tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $\mu$ of $m(x)$, satisfying as it does $\mu \ne 0$, is a unit; thus $\mu^{-1} m(x)$ is monic and
$(\mu^{-1} m(x)) = (m(x)); \tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), \; p(x), q(x) \in F[x], \; \deg p(x), \deg q(x) \ge 1; \tag 5$
consider the ideal
$(p(x)) \subsetneq F[x]; \tag 6$
it is clearly proper: since $\deg p(x) \ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$ other than $0$. Also,
$(m(x)) = (p(x)q(x)) \subsetneq (p(x)), \tag 7$
for
$p(x) \notin (p(x)q(x)) \tag 8$
lest for some
$r(x) \in F[x] \tag 9$
we have
$p(x) = r(x)p(x)q(x), \tag{10}$
or
$p(x)(r(x)q(x) - 1) = 0, \tag{11}$
whence
$r(x)q(x) = 1, \tag{12}$
which yields
$\deg r(x) + \deg q(x) = \deg 1 = 0, \tag{13}$
impossible in light of the assumption $\deg q(x) \ge 1$; thus we have shown that
$(m(x)) = (p(x)q(x)) \subsetneq (p(x)) \subsetneq F[x] \tag{14}$
in the event that $m(x)$ is reducible, which further shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.
Best Answer
Since every odd degree real polynomial has a root in $\mathbb{R},$ an irreducible polynomial of degree $> 2$ must be of even degree. Let $p(x) \in \mathbb{R}[x]$ be an irreducuble polynomial of degree $2d, d \geq 1.$ We can also assume that it is monic. In $\mathbb{C}[x], p(x)$ will factor in linear factors. Let $p(x) = \prod_{1 \leq i \leq2d} (x - \alpha_i) \in \mathbb{C}[x].$ Since complex roots occurs in conjugate pairs, we can rename the $\alpha_i$'s to write $p(x) = \prod_{1 \leq i \leq d} (x - \alpha_i)(x - \overline{\alpha}_i)$ where $\overline{\alpha}_i$ is the complex conjugate of $\alpha_i.$ From this we can write, $p(x) = \prod_{1 \leq i \leq d}q_i(x)$ where $q_i(x) = (x - \alpha_i)(x - \overline{\alpha}_i)$ is a real quadric polynomial. This shows that $d = 1.$ So $p(x)$ is of the form $ax^2 + bx + c$ with $b^2 - 4ac < 0.$
Hence the maximal ideals of $\mathbb{R}[x]$ are of the form $(ax^2 + bx + c)$ with $b^2 - 4ac < 0$ or of the form $(x - a), a \in \mathbb{R}.$