= Solution
Write $\mathfrak p=(x)$. The principal ideal theorem and the hypothesis give $\operatorname{ht}\mathfrak p=1$. We use the standard <principal prime in a Noetherian local ring lemma>: a principal prime of positive height in a Noetherian local ring is generated by a nonzerodivisor and is the unique minimal prime above zero. The lemma follows by applying the associated-prime description of zero divisors and Nakayama's lemma to $\operatorname{ann}(x)$; if a nonzero annihilator or another minimal component existed, the principal prime would have height zero.
Here is the needed argument directly. For every $n\geq1$,
$$
\operatorname{ann}(x^n)\subseteq\mathfrak p.
$$
Indeed, if $yx^n=0$ with $y\notin\mathfrak p$, then $y$ is a unit in $R_{\mathfrak p}$, so $x^n=0$ there. The maximal ideal $\mathfrak pR_{\mathfrak p}=(x)$ would then be nilpotent, making $R_{\mathfrak p}$ zero-dimensional, contrary to $\operatorname{ht}\mathfrak p\geq1$.
Now suppose $xy=0$. The displayed containment gives $y=xy_1$, then $x^2y_1=0$ gives $y_1=xy_2$, and inductively
$$
y=x^ny_n,\qquad x^{n+1}y_n=0.
$$
Because $R$ is Noetherian, the ascending chain $\operatorname{ann}(x)\subseteq\operatorname{ann}(x^2)\subseteq\cdots$ stabilizes, say at $n=N$. Then $y_N\in\operatorname{ann}(x^{N+1})=\operatorname{ann}(x^N)$, and hence $y=x^Ny_N=0$. Thus $x$ is a nonzerodivisor.
Let $N=\bigcap_{n\geq0}x^nR$. It is a finitely generated ideal. If $y=xz\in N$, then $xz\in x^{n+1}R$ for every $n$; cancellation of the nonzerodivisor $x$ gives $z\in x^nR$ for every $n$. Hence $N=xN$, and <Nakayama lemma> gives $N=0$ because $x$ lies in the maximal ideal.
Every nonzero element consequently has a finite $x$-adic order. If nonzero $a,b$ satisfied $ab=0$, write $a=x^ra'$ and $b=x^sb'$ with $a',b'\notin(x)$. Cancelling $x^{r+s}$ gives $a'b'=0$, contradicting primality of $(x)$. Equivalently, $(0)$ is prime, so
$$
\boxed{R\text{ is an integral domain}.}
$$
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