Every maximal ideal is prime in a commutative ring with identity.
There were several posts on the site about analogues of the claim for rngs (rings with or without identity):
A maximal ideal is always a prime ideal?
and
Examples of a commutative ring without an identity in which a maximal ideal is not a prime ideal
It looks like they do not provide the correct extension of the claim onto rngs.
We call an ideal maximal if it is a maximal proper ideal in the poset of ideals.
This notion assumes that the only "bigger" ideal for a maximal ideal is the principal ideal of units in a ring with identity.
It looks like the correct extension of the notion of a maximal ideal onto rngs is not a maximal proper ideal, but a maximal non-unit ideal (a maximal ideal in the poset of ideals that are not generated by units).
For example, the ideal $2 \mathbb Z$ is a maximal non-unit ideal in the ring with identity $\mathbb Z$, and it is prime;
the ideal $2 \mathbb Z$ is a maximal non-unit ideal in the ring without identity $2 \mathbb Z$, and it is prime.
In this case the claim for maximal ideals in rngs should be formulated in the following way:
every maximal non-unit ideal is prime in a commutative rng.
Is this correct?
Is there any use of the term "maximal non-unit ideal"?
It all looks standard to me.
If a ring does not have an identity, then it does not have units either (the definition of a unit requires the existence of an identity.) So, the proposed "better" definition for maximal ideals in rings does not have any meaning in a ring without identity.
One can argue, however, that the definition of a maximal ideal (for rings with identity) should be elaborated on to make it work in rings without identity.
One way to do this, as Jacobson did, is to additionally require the ideal to be modular. To accurately state it, he called a right ideal $T$ of $R$ modular if there exists an element $e\in R$ such that $ex=x$ for all $x\in T$. Put another way, there is an element that acts like a left identity on $T$. Notice how when a ring has identity, $e=1$ works for all right ideals maximal in the poset of proper right ideals, so they are all modular. This is a "good" extension of the "absolute" definition of maximal right ideals.
He used these ideals to characterize the Jacobson radical of rings without identity as the intersection of maximal modular right ideals (and not the "absolutely" maximal right ideals.)
In the most common example given in the posts you linked, the rng in question is $R=2\mathbb Z/4\mathbb Z$. Now, the zero ideal is certainly a maximal proper ideal in the ring, but it fails to be modular, as you can see. For this reason, $J(R)=R$, and not the zero ideal.