Answering my own question. I decided to take a look at Anderson and Valdes-Leon's paper (1996).
They do consider this notion, and they call it m-irreducible (p. 448):
A nonunit $a \in R$ is m-irreducible if $(a)$ is a maximal element in the set of proper ideals of $R$.
This notion is related to three other notions of irreducibility, none of them equivalent (see also the discussion at this MO answer). Here is a diagram of all the relevant notions ordered by inclusion (most of these I know are strict, but I may be missing some inclusions):
maximal very strongly irreducible
| \ /
| m-irreducible
prime |
| strongly irreducible
| |
weakly irreducible
Maximal and prime are standard ($p$ is prime if $(p)$ is a prime ideal and $p$ is maximal if $(p)$ is a maximal ideal). The rest of the categories above can be restated as follows, in decreasing order of strength. I use weakly irreducible for what they call irreducible to avoid confusion with the usual definition.
$p$ is very strongly irreducible if it is a nonunit and whenever $p = ab$, either $a$ or $b$ is a unit. (Note: this is the usual definition of irreducibility that I am familiar with.)
$p$ is m-irreducible if it is a nonunit and whenever $p = ab$, $(a) = (1)$ or $(a) = (p)$. (Note: this is the definition of minimal under division in the OP.)
$p$ is strongly irreducible if it is a nonunit and whenever $p = ab$, there exists a unit $u$ such that $p = ua$ or $p = ub$.
$p$ is weakly irreducible if it is a nonunit and whenever $p = ab$, we have $(a) = (p)$ or $(b) = (p)$. (Note: this is similar to, but weaker than $p$ being prime.)
The following is adapted from Anderson and Valdes-Leon (Theorem 2.13, p. 450):
Theorem 2.13. Let $a \in R$. Then $a$ very strongly irreducible $\implies$ $a$ is $m$-reducible $\implies$ $a$ is strongly irreducible $\implies$ $a$ is weakly irreducible.
Moreover, none of these implications can be reversed.
Proof.
We show each of the three inclusions in turn (the second inclusion is the most interesting).
First, $p$ is very strongly irreducible. For any decomposition $p = ab$, either $a$ is a unit or $b$ is a unit. In the former case, $(a) = (1)$. In the latter case, $(b) = (1)$, so $(a) = (a)(1) = (a)(b) = (p)$. Thus, $p$ is m-irreducible.
Next, suppose $p$ is m-irreducible. For any decomposition $p = ab$, if either $(a) = (1)$ or $(b) = (1)$ then we are done as $p$ is equal to one of $a$ or $b$ times a unit. Thus it remains to consider $(a) = (b) = (p)$. In this case, we can write $a = pa'$ and $b = pb'$, so $p = ab = p^2 a' b'$. Letting $r = a' b'$, we have $p = p^2r$ so $pr = (pr)^2$, i.e. $pr$ is idempotent. Also, $(pr) = (p)$. Let $e = pr$ be the idempotent.
Now we claim that, for any $x$, if $(x) = (e)$ then $x = eu$ for some unit u. For this, consider the quantity
$$
u = ex + (1-e).
$$
Multiplying by $e$ we have $eu = ex + 0 = ex = x$ since $e$ divides $x$. But this is also a unit: since $x$ divides $e$ we have $e = xy$, then letting $v = ey + (1-e)$ we get
$$
uv = exy + (1-e)^2 = e + 1-e = 1,
$$
hence $x = eu$ and $u$ is a unit.
Applying this result both $x = a$ and $x = p$, we get that $a = eu_1$ and $p = eu_2$, hence $a$ and $p$ are equivalent up to multiplication by a unit. So $p$ is strongly irreducible.
Finally, suppose $p$ is strongly irreducible. For any decomposition $p = ab$, suppose WLOG $p = ua$. Then $(p) = (a)$. Thus, $p$ is weakly irreducible.
Notes
Anderson and Valdes-Leon also say in Theorem 2.13 that $p$ is strongly irreducible implies that $a$ is both weakly irreducible and prime. But this seems like a mistake in the text; they don't prove it, and it's false for integral domains. That is, it is well-known that there are examples of irreducible (hence very strongly irreducible) elements that are not prime; see this question. The rest of the theorem seems OK.
Anderson and Valdes-Leon derive weakly, strongly, and very strongly irreducible definitions from corresponding relations $\sim$, $\approx$, $\approxeq$ (associates, strong associates, and very strong associates) on $R$. Above I've given direct definitions (using some of the equivalent statements) without using $\sim$, $\approx$, and $\approxeq$.
My definition of "very strongly irreducible" differs from Anderson and Valdes-Leon in case $p = 0$. In that case, they decide to special case $0 \approxeq 0$, and it follows that $0$ is very strongly irreducible precisely when $R$ is a domain. But this later becomes awkward, as it requires them to exclude $a \ne 0$ in Theorem 2.13. With the above definitions the theorem holds for $a = 0$: $0$ is never very strongly irreducible; it's $m$-reducible exactly when $R$ is a field; and it's strongly and weakly irreducible precisely when $R$ is a domain.
