If $R$ is a ring, possibly without $1$, a right ideal $\mathfrak{a}$ of $R$ is modular if there exists $e\in R$ such that $r-er\in \mathfrak{a}$ for all $r\in R$. So $e$ is a left identity mod $\mathfrak{a}$.
In Lam's book A First Course In Noncommtative Rings, exercise 4.5(a) says that every proper modular right ideal $\mathfrak{a}$ of $R$ is contained in a modular maximal right ideal $\mathfrak{m}$. (To be clear: $\mathfrak{m}$ should be a maximal right ideal which just happens to be modular.)
I also found the following quote in the wikipedia article for regular ideal:
Somewhat surprisingly, it is possible to prove that even in rings without identity, a modular right ideal is contained in a maximal right ideal.
But this article doesn't specify that the maximal ideal is modular.
How do you solve Lam's exercise? I've tried Zorn's Lemma: if $Z$ is the set of proper modular right ideals containing $\mathfrak{a}$, then $Z$ is nonempty since $\mathfrak{a}\in Z$. Suppose $\{\mathfrak{b}_i\}$ is a chain in $Z$. Probably a candidate for an upper bound is the union $\mathfrak{b}:=\bigcup \mathfrak{b}_i$. This is a right ideal, but I'm not sure why it's proper or modular though. In the unital case one can conclude $\mathfrak{b}\subsetneq R$ because $1\notin \mathfrak{b}$. But I don't know how to adapt this to the nonunital case.
Is this the wrong approach?
If $\mathfrak{a}$ is a modular right ideal, then every right ideal $\mathfrak{b}$ containing $\mathfrak{a}$ is modular as well, because $r-er\in\mathfrak{a}$ implies $r-er\in\mathfrak{b}$.
Note also that $e\notin\mathfrak{a}$, if $\mathfrak{a}$ is a proper ideal.
Applying Zorn's lemma is indeed the way to go. Consider the family of right ideals containing $\mathfrak{a}$ and not containing $e$, ordered by inclusion.
The union of a chain in this family is a (modular) right ideal containing $\mathfrak{a}$ and not containing $e$; so Zorn's lemma provides a maximal member of the family, $\mathfrak{m}$.
Suppose a right ideal $\mathfrak{b}$ properly contains $\mathfrak{m}$; then $\mathfrak{b}$ contains $\mathfrak{a}$ and, by maximality of $\mathfrak{m}$ it must also contain $e$. Therefore $\mathfrak{b}=R$.