There are indeed very many rings in which the nilradical equals the Jacobson radical.
Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.
And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:
Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.
Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.
For some information on this subject, including a proof of the theorem, see these notes.
Claim: $A$ is a ring such that every primary ideal is prime if and
only if $A$ is absolutely flat.
Let us say a ring is a PP(primary is prime) ring if every primary ideal is
prime.
Suppose $A$ is absolutely flat, then $A$ is PP. (exercise 3 in page 55 of "Introduction to Commutative Algebra" by Atiyah & Macdonald)
We only need to show that if $A$ is PP then $A$ is absolutely flat.
Notice that every primary ideal of $S^{-1}A$ is of the form $S^{-1}\mathfrak{q}$
where $\mathfrak{q}$ is a $\mathfrak{p}$-primary ideal such that
$\mathfrak{p}\cap S=\emptyset$. And if $J/I$ (here $J\supset I$) is a
primary ideal of $A/I$ then every zerodivisor in $A/J=(A/I)/(J/I)$ is
nilpotent, so $J$ is is primary ideal of $A$.
Now it is easy to show that, if $A$ is PP, then $S^{-1}A$ and
$A/I$ is PP for a multiplicatively closed subset $S$ and an ideal
$I$ and $\mathfrak{m}^2=\mathfrak{m}$ for every maximal ideal
$\mathfrak{m}$.
Based on these properties we are able to prove that PP rings are absolutely
flat.
We first show every prime ideal in $A$ is maximal. Suppose not, there are
two distinct prime ideals $\mathfrak{p}\subset \mathfrak{m}$ where
$\mathfrak{m}$ is maximal. Now $B=(A/\mathfrak{p})_{\mathfrak{m}}$ is
a PP, local domain(but not a field) we also use $\mathfrak{m}$ to
denote the maximal ideal of $B$. Let $0\neq b\in \mathfrak{m}$,
suppose $\mathfrak{q}$ is a minimal prime containing $b$, then
$C=B_{\mathfrak{q}}$ is a PP, local domain(not a field) and the only maximal ideal of
$C$ is minimal over the ideal $(b)=bC$ so $(b)$ is primary and hence
maximal, $(bC)^2=bC$, thus $bC=0$. It is impossible. We have proved
that every prime ideal in $A$ is maximal.
Let $\mathfrak{m}$ be any prime ideal in $A$, then $A_{\mathfrak{m}}$
is PP. If $A_{\mathfrak{m}}$ is not a field, pick any nonezero $x\in
\mathfrak{m}A_{\mathfrak{m}}$, then $(x)$ is primary hence equals
$\mathfrak{m}A_{\mathfrak{m}}$, so $\mathfrak{m}A_m=0$,
contradiction.
Hence every prime ideal $\mathfrak{m}$ in $A$ is maximal and
$A_{\mathfrak{m}}$ is a field, so $A$ is absolutely flat. We are done.
Best Answer
You won't find any examples of maximal, non-prime ideals other than those given in Arturo Magidin's lovely answer. I won't even assume commutativity. And I freely admit that this is basically the same argument as in Arturo's answer!
Claim: If $R$ is a rng with a maximal ideal $M$ that is not prime, then $R^2 \subseteq M$.
Proof: Let $M$ be such an ideal, and suppose that $A,B$ are ideals of $R$ not contained in $M$ such that $AB \subseteq M$. By maximality of $M$ we have $M + A = R = M + B$. It follows that $$\begin{align*} R^2 &= (M+A)(M+B) \\ &= M^2 + AM + MB + AB \\ &\subseteq M, \end{align*}$$ since $M^2,AM,MB,AB \subseteq M$. QED
Combining this with Arturo's theorem, we have:
Corollary: Let $R$ be a rng with maximal ideal $M$. Then $M$ is not prime if and only if $R^2 \subseteq M$.