$\pi$ is irreducible iff $\langle\pi\rangle$ is a maximal ideal?

Question

Let $R$ be a domain of principal ideals. Then prove that $\pi \in R$ is irreducible $\Leftrightarrow$ $\langle\pi\rangle$ is a maximal ideal. Now I don't remember if R has multiplicative identity in the hypothesis, but I think it must have it. Also, here I use "ideal"$\equiv$"bilateral ideal". I've already proved the first part, so I have $\pi$ irreducible $\Rightarrow\ \langle\pi\rangle$ is maximal, but i have trouble with the second part. So suppose $\langle\pi\rangle$ is maximal $\Rightarrow\ \langle\pi\rangle$ is prime, i.e. $\forall\ a,b\in R$ such that $ab\in \pi R=R\pi=\langle\pi\rangle$ (since R is an IPD), then $a\in \pi R$ or $b\in \pi R$, this implies $\pi\mid a$ or $\pi\mid b\ $ i.e. $a=\pi\cdot r'$ or $\ b=\pi\cdot r''$ and $ab=\pi\cdot r$ for some $r,r',r''\in R$ but I don't see how this shows that $\pi$'s only divisors are $\pi$ itself and unities (here's why I say $R$ must have $1$, since all unities are asociated to $1$ [I'm using "$1$"$\equiv$"multiplicative identity"]).

Answer 1

Hint:

Let $\pi=fg$ where $f,g\in R$, then $\left<\pi\right>\subseteq\dots$

Answer 2

Since you wrote as if you were considering noncommutative PID's (you emphasized you were talking about bilateral ideals, and you used $\langle\cdot \rangle$) I offer these comments.

It happens that there exist simple left-and-right PID's which aren't division rings. In such a ring, $\langle x\rangle$ is the entire ring for every nonzero element, even "irreducibles" (however you are defining them).

In such a ring, it is not necessarily true that any of $xR, Rx$ or $\langle x\rangle$ are equal, although the lattermost contains the first two, of course.

Along with being more careful about defining irreducibility in a noncommutative PID, you'd have to be careful defining divisibility as well. It is pretty rare to see the noncommutative version discussed, although I know several papers by P. M. Cohn apply.

For these reasons, I think you're better off confining yourself to the commutative version so that you can follow your line of thought to a successful conclusion.