To address your last question: It is not true in general that maximal ideals are necessarily prime in (commutative) rings without identity.
Consider the ring without identity $R=2\mathbb{Z}$. The ideal $4\mathbb{Z}$ is maximal in $R$ (since it has prime index as a subgroup), but it is not prime: $2\times 2\in 4\mathbb{Z}$, but $2\notin 4\mathbb{Z}$.
On the other hand,
Proposition. Let $R$ be a ring, not necessarily commutative, not necessarily with identity, such that $RR=R$ (in particular, this holds if $R$ has an identity). If $\mathfrak{M}$ is a maximal ideal, then $\mathfrak{M}$ is a prime ideal; that is, if $\mathfrak{AB}\subseteq \mathfrak{M}$ for ideals $\mathfrak{A}$ and $\mathfrak{B}$, then $\mathfrak{A}\subseteq\mathfrak{M}$ or $\mathfrak{B}\subseteq \mathfrak{M}$.
Proof. Let $\mathfrak{M}$ is a maximal ideal of $R$, and $\mathfrak{A},\mathfrak{B}$ are ideals such that neither $\mathfrak{A}$ nor $\mathfrak{B}$ are contained in $\mathfrak{M}$; we will show that $\mathfrak{AB}$ is not contained in $\mathfrak{M}$. Indeed, maximality of $\mathfrak{M}$ gives $\mathfrak{A}+\mathfrak{M}=\mathfrak{B}+\mathfrak{M} = R$, so $R = RR = (\mathfrak{A}+\mathfrak{M})(\mathfrak{B}+\mathfrak{M}) = \mathfrak{AB}+\mathfrak{AM}+\mathfrak{MB}+\mathfrak{MM}\subseteq \mathfrak{AB}+\mathfrak{M}\subseteq R,$ so $\mathfrak{AB}+\mathfrak{M}=R$, hence $\mathfrak{AB}$ is not contained in $\mathfrak{M}$. $\Box$
The condition that $RR=R$ is a bit tricky. There are rings in which this does not hold but the implication holds anyway: for example, take an abelian group that has no maximal ideals (e.g., $\mathbb{Q}$), and make it into a ring by defining $ab=0$ for all $a,b$. Then ideals are subgroups, and the absence of maximal ideals means that the implication holds by vacuity. If $RR\neq R$ and there is a maximal ideal that contains $RR$, then that ideal will be a witness to the implication not holding, as occurs for example above with $R=2\mathbb{Z}$.