In this case there is an easier direct argument without using Krull's intersection theorem. First, note that by Nakayama's lemma, if $p\in\mathfrak{m}\setminus\mathfrak{m}^2$, then $p$ generates $\mathfrak{m}$ (since $\mathfrak{m}/\mathfrak{m}^2$ is $1$-dimensional and hence generated by any nonzero element). It follows that $\mathfrak{m}^n$ is generated by $p^n$, and $\bigcap_{n>0} \mathfrak{m}^n$ is the set of elements which are divisible by $p^n$ for all $n$.
Now suppose $x\in \bigcap_{n>0} \mathfrak{m}^n$ is nonzero and let $x_n$ be such that $p^nx_n=x$ for each $n$. I claim that for each $n$, $x_n\not\in(x_{n-1},\dots,x_0)$, so that the ideals $(x_0)\subset(x_0,x_1)\subset(x_0,x_1,x_2)\subset\dots$ are an infinite ascending chain and contradict the assumption that $R$ is Noetherian. To prove this, suppose $x_n\in (x_{n-1},\dots,x_0)$ and write $x_n=\sum_{i=0}^{n-1} a_ix_i$ for $a_i\in R$. Multiplying by $p^n$, we get $x=\sum a_ip^{n-i}x$, so $(1-\sum a_ip^{n-i})x=0$. But $\sum a_ip^{n-i}$ is divisible by $p$ (since $i$ stops at $n-1$) and hence in $\mathfrak{m}$ so $1-\sum a_ip^{n-i}$ is a unit. This contradicts the assumption that $x$ is nonzero.
(If you additionally assume $R$ is a domain, which seems to be intended in your context (otherwise the conclusion that $R$ is a DVR is not correct!), there is an even quicker way to formulate this proof. Namely, in that case if $x$ is infinitely divisible by $p$ there is a unique way to divide it by $p$, which means $x/p$ is also infinitely divisible by $p$. That is, for any $x\in\bigcap_{n>0} \mathfrak{m}^n$, $x/p\in \bigcap_{n>0} \mathfrak{m}^n$ as well, which means $p\cdot \bigcap_{n>0} \mathfrak{m}^n=\bigcap_{n>0} \mathfrak{m}^n$. By Nakayama, this implies $\bigcap_{n>0} \mathfrak{m}^n=0$.)
