Commutative rings without assuming identity

You don't need an identity. Let $P$ be a prime ideal and let $I$ be an ideal which strictly contains $P$. I will show that $I$ is the whole ring.

Let $i \in I - P$. For any element $r$ in the ring, $$ i(r-ri) = ri-ri^2 = ri - ri = 0 \in P. $$ Since $P$ is a prime ideal, this implies that $r-ri$ is in $P$. Since $ri \in I$, this shows that $r = (r-ri)+ri \in I$.

Neither do you need to assume that the ring is unital nor the ring is commutative.

Definition: Those rings where for each $x \in R$ there is a positive integer $n(x) >1$ (depending on $x$) s.t. $x^{n(x)}=x,$ are called Jacobson-rings or J-rings.

General result due to Jacobson: J-rings are commutative. For the proof, you may take a look at Non-commutative Rings by Herstein)

In any J-ring, every prime ideal is maximal.

Proof: Let $p$ be a prime ideal which is not maximal. So there is a maximal ideal $m$ of $R$ s.t. $p \subsetneq m \subsetneq R.$ Let $x \in m\setminus p.$ By the given property, there is a natural number $n>1$ s.t. $x^n=x$ or $x^n-x=0.$ Let $y \in R$ be arbitrary. Then $x^ny-xy=0 \to x(x^{n-1}y-y)=0 \in p$, therefore, either $x \in p$ or $x^{n-1}y-y \in p.$ Now $x \in m\setminus p$, we conclude that $x^{n-1}y-y \in p \subsetneq m.$ We know that $x \in m$ therefore, $y \in m.$ Since $y$ was an arbitrary element of $R$ then $m=R$ which is a contradiction and we're done.

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 $R=2\mathbb{Z}$.