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 $R^2 \subseteq M$.
Wow, three answers and a comment with the same example, and not even the easiest example, IMO! (It's still a good example, though!)
Similarly, is there an example within the quaternions?
Absolutely! There are lots of nonzero subrings of $\Bbb H$ that would already work. Consider for example $\{a+bi+cj+dk\mid a,b,c,d\in \Bbb Z\}\subseteq\Bbb H$.
It's easy to check that this is a subring of $\Bbb H$, which automatically makes it a domain. Clearly it isn't commutative since $ij\neq ji$. Finally, it doesn't contain $2^{-1}$, so it can't be a division ring.
Is there an example within some ring of matrices?
Absolutely! Actually this will be cheating since it will be the same as my first solution, but you might be interested in what happens anyway.
It turns out that the quaternions can be represented by complex matrices in this way:
$a+bi+cj+dk\mapsto\begin{bmatrix}a+bi&c+di\\-c+di&a-bi\end{bmatrix}\in M_2(\Bbb C)$. So by just taking the image of the ring I described in the first part, you have a subring of $M_2(\Bbb C)$ which is a noncommutative domain, not a division ring.
You could take it even further to real matrices:
$a+bi+cj+dk\mapsto\begin{bmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\\\end{bmatrix}\in M_4(\Bbb R)$ to get another (the same) example. In fact, you can look upon the final example as using a matrix representation of elements of $\Bbb C$: $a+bi\mapsto\begin{bmatrix}a&b\\-b&a\end{bmatrix}$ and plugging that representation into the second example. All three are interlinked.
These are both matrix representations of the original example I suggested.
For a final exotic example that you might be interested in, check out the Hurwitz quaternions. They are an interesting subdomain of $\Bbb H$.
Best Answer
In other words you are looking for a ring with unique maximum ideal that is not simple and not local.
Take any noncommutative simple ring $R$ that is not a division ring. You said you knew a couple: you could take the two by two matrix ring over the reals or complex numbers.
Form the trivial extension $S=R\times R$ where the addition operation is coordinatewise and multiplication is $(a,b)(c,d)=(ac,ad+bc)$.
Since $I=\{0\}\times R$ is a nonzero nilpotent ideal, $S$ is not simple. Furthermore the maximal ideals of $S$ correspond to those of $S/I$ which is simple, so $I$ is the unique maximal ideal, but not the unique maximal right ideal (since $S/I$ lacks a unique maximal right ideal.)