Examples of a commutative ring without an identity in which a maximal ideal is not a prime ideal In a commutative ring with an identity, every maximal ideal is a prime ideal.
However, if a commutative ring does not have an identity, I'm not sure this is true.
I would like to know the counterexamples, if any.
The more examples, the better.
EDIT
I would like to know the counterexamples other than $2\mathbb{Z}$.
The more examples, the better.
EDIT
I also would like to know the counterexamples that are not given in the Arturo Magidin's answer if any, namely an example of a non-prime maximal ideal which does not contain $R^2$. 
 A: There are lots of parodies on the $2\Bbb Z$ example.
You can look at $R=2\Bbb Z[x]$: the ideal $(2x,4)=I$ is maximal, and again $2^2\in I$ and $2\notin I$.
Here's another: let $M$ be any $2\Bbb Z$ module. You could, for example, let $M=2\Bbb Z$, or any number of copies of $2\Bbb Z$. Look at the following ring of matrices and ideal:
$$
R=\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\mid a\in 2\Bbb Z, b\in M\right\}
$$
$$
I=\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\mid a\in 4\Bbb Z, b\in M\right\}
$$
Again, $\begin{bmatrix}2&0\\0&2\end{bmatrix}^2$ shows $I$ isn't prime.
Finally to get a noncommutative example, try
$$
R=\begin{bmatrix}2\Bbb Z&2\Bbb Z\\0&2\Bbb Z\end{bmatrix}
$$
$$
I=\begin{bmatrix}2\Bbb Z&2\Bbb Z\\0&4\Bbb Z\end{bmatrix}
$$
$$
J=\begin{bmatrix}4\Bbb Z&2\Bbb Z\\0&2\Bbb Z\end{bmatrix}
$$
$I$ is a maximal ideal of $R$, $J\nsubseteq I$, but $J^2\subseteq I$, so $I$ isn't prime.
A: I will prove the following fact.
Let $R$ be a commutative ring without an identity.
Let $M$ be a non-prime maximal ideal.
Then $R/M$ has a prime order.
Proof
Let $S = R/M$.
Since $R^2 \subset M$ by the Manny Reyes' answer, $S^2 = 0$.
Hence every subgroup of the additive group $S$ is an ideal of $S$.
Since $S$ has no non-trivial ideals, the order of $S$ must be a prime number.
QED
A: 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 $R^2 \subseteq M$.
A: Take $R=2\mathbb{Z}$, the ring of even integers. The ideal $4\mathbb{Z}\subset R$ is maximal (the only larger ideal is $R$ itself), but not prime, as $2\cdot 2\in 4\mathbb{Z}$, but $2\not\in 4\mathbb{Z}$.
A: Let $R$ be an abelian group of prime order.
We define multiplication on $R$ by $ab = 0$ for all $a, b \in R$.
Then $0$ is a maximal ideal but not a prime ideal.
A: The following example is a generalization of the example $2\mathbb{Z}$.
Let $p$ be a prime number.
Let $R = p\mathbb{Z}$.
Then $R^2 = p^2\mathbb{Z}$ is a maximal ideal but not a prime ideal of $R$.
