= Solution
Since the <irreducible element> $x$ is not a <unit>, the <principal ideal> $(x)$ is proper. Suppose $(x)\subseteq J\subseteq R$. As $R$ is a <principal ideal domain>, write $J=(d)$. Then $x=dr$ for some $r\in R$. The definition of an <irreducible element> says that $d$ or $r$ is a <unit>.
If $d$ is a <unit>, then $J=R$. If $r$ is a <unit>, then $d$ and $x$ are <associate elements>, so $J=(x)$. Thus $(x)$ is a <maximal ideal>. By the permitted criterion that a <quotient ring> by a <maximal ideal> is a <field>,
$$
\boxed{R/(x)\text{ is a field}}.
$$
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