$(R,m)$ is a noetherian local ring. Suppose $\dim(R)=1$ and $m$ principal. Does it follow that $R$ is a domain?
"A local ring $(R,m)$ which is not a field is a valuation ring iff $\mathrm{ht}(m)\geq 1$ and $m$ is principal." (Note that the book assumes that local ring $(R,m)$ is noetherian local ring.)
Ref. Nagata, Local Rings, Chpt. 1, Sec. 12
On
The hypothesis that $R$ is local and Noetherian implies that $\bigcap \mathfrak{m}^n = 0$ (argue e.g. by Krull intersection theorem and Nakayama's lemma).
That $\mathfrak{m} = mR$ is principal then implies that every element of $R$ can be written uniquely as a power of $m$ times a unit.
It follows that $R$ has a zero-divisor iff $m$ is nilpotent.
But if $m$ is nilpotent, then $R$ is $0$-dimensional. Conclude that
A local Noetherian ring $(R, \mathfrak{m})$ with $\mathfrak{m}$ principal is a Principal Ideal Ring which is either:
$(1)$ a ring with exactly one prime ideal which is nilpotent
$(2)$ a $1$-dimensional domain.
Let $x$ be a generator of $m$ and let $\mathfrak{p}$ be any minimal prime ideal of $R$. Then $\mathfrak{p} \subset xR$.
So if $u \in \mathfrak{p}$, we can write $u=xv$ for some $v \in R$. Since $m$ is not minimal, $x \notin \mathfrak{p}$, so $v \in \mathfrak{p}$. As a consequence, $\mathfrak{p} \subset m\mathfrak{p}$, yielding $\mathfrak{p}=0$ by Nakayama, thus $0$ is a prime ideal and $R$ is a domain.