 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,