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 $R^2 \subseteq M$.
I think you're blurring two notions of "radical": the radical of an ideal and the radical of a ring.
A radical of an ideal $I$ is often defined by taking the intersection of special ideals containing $I$.
The radical of a ring is often defined from a radical of an ideal by saying "The radical of $R$ is defined to be the radical of the zero ideal in $R$."
So: in your first paragraph you are looking at a radical of an ideal, and in your second statement, you are thinking of a radical of a $I$ considered as a ring.
Another illustration for commutative rings is $\sqrt{I}:=\cap\{P\mid P\text{ prime and } I\subseteq P\}$. This is called the radical of $I$, and the radical of the ring $R$ is $\sqrt{0}$, the radical (nilradical, more commonly) of the ring.
You can do something analogous with modules and submodules.
Generally speaking, though, the notation $J(-)$ is not standardly employed for radicals of ideals, but it is usually used radicals of rings/modules. If you use $J(-)$ to denote a radical of an ideal with the definition you used, then $J(R)=R$ in every case, because the set of maximal ideals containing $R$ is empty, and the empty intersection of ideals is usually interpreted as the whole ring. More standardly, $J(R)$ denotes the intersection of maximal ideals containing $\{0\}$, and this can be equal to all of $R$ in rings without identity. It's not $R$ if $R$ has identity, though.
Best Answer
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.