If $R[x]$ is a PID, $R$ necessarily has to be a field?

Question

It is given that R is a commutative ring with identity.

My attempt: I tried to get a contradiction. Given a nonunit $a \in R$, I wanted to show that $(a,x)$ is not a principal ideal in $R[x]$. Then, by being PID, $(a,x)=(b(x))$ where $ b(x) \in R[x]$. But I do not know how to go on from here.

Answer 1

In a PID, every prime ideal is maximal. Here $(x)$ is a prime ideal and hence a maximal ideal. Thus $R[x]/(x)$ is a field. But this field is nothing but....?