Theorem 5.2 in http://www.emis.de/journals/BAG/vol.46/no.1/b46h1her.pdf gives an answer
(take the projective limit. The paper has a related one with corrections,
but not for the part that is related to your question). This is for a non-commutative
case, and the theorem has a non-commutative extension: a PIR is a finite direct product
of prime and artinian indecomposable cases, which are matrix rings over CPU rings
(Faith, Algebra II should contain all the needed references)
Even in the noncommutative case, we can show that a local ring whose maximal ideal $\mathfrak{m}$ is left principal is a left PIR if and only if $\bigcap_{i \geq 0} \mathfrak{m}^i = 0$ (the conclusion of Krull's intersection theorem).
Recall that a noncommutative ring $R$ is local if the non-units form a left ideal $\mathfrak{m}$ (and in this case $\mathfrak{m}$ is in fact a two-sided ideal). Suppose this maximal ideal is left-principal, say $\mathfrak{m} = Rt$.
The "if" direction follows the same proof as in the commutative case.
For the "only if" direction, suppose that $R$ is a left PIR. Let $I = \bigcap_{i \geq 0} \mathfrak{m}^i$. Since $R$ is a PIR, we have $I = Rx$ for some $x$. Similar to the proof of Krull's intersection theorem in the commutative case, show that $tI =I$ (if $z \in I$, then $z = tz'$ for some $z'$; if $z' \notin I$, then $z' \notin \mathfrak{m}^j$ for some $j$, but then $z \notin \mathfrak{m}^{j+1}$, contradicting $z \in I$). Then there is some $y \in I$ such that $x = ty$. Since $y \in I=Rx$, we have $y=rx$ for some $r$, and thus $x = trx$, or $(1-tr)x=0$. Since $tr \in \mathfrak{m}$, $1-tr$ is a unit, and thus $x=0$ and $I = Rx = 0$.
(I don't even think you need Noetherian to prove this; in case $R$ ends up being a left PIR, Noetherian seems to follow.)
Best Answer
Here is an elementary argument. Let $R$ be a local ring with a principal maximal ideal $M=mR$.
Since $1+M\subseteq R^*$, we have $ma\mid a$ only if $a=0$. Moreover, every nonunit is divisible by $m$, hence if there is a nonzero $a\in\bigcap_nM^n$, we can construct an increasing sequence of ideals $a_0R\subsetneq a_1R\subsetneq a_2R\subsetneq\cdots$ where $a_0=a$ and $ma_{n+1}=a_n$, in particular $R$ is not noetherian.
On the other hand, if $\bigcap_nM^n=0$, every nonzero element can be written as $um^n$ for some $u\in R^*$ and $n\in\omega$, so all nonzero ideals are of the form $M^n$. This makes $R$ a discrete valuation ring unless $m$ is nilpotent. Thus,
Fact: If $R$ is a local ring with a principal maximal ideal $M$, the following are equivalent.
$R$ is noetherian.
$\bigcap_{n\in\omega}M^n=0$.
$R$ is a DVR, or there is $n\in\omega$ such that all ideals of $R$ are $R,M,M^2,\dots,M^n=0$.