Let $R$ be an integral domain which is not a field. Then is it true that $\langle f \rangle$ cannot be a maximal ideal of $R[x]$ for any non-constant polynomial $f(x) \in R[x]$ ?
I know that it holds in case of $R=\mathbb Z$ and I can adapt the proof to draw similar conclusion in case $R$ is a UFD with infinitely many mutually non-associated prime elements. But I don't know what happens in general. If the answer in general is not true then does it at least hold if we also assume $R$ is Noetherian ?
Let $K$ be a field, and let $R=K[[y]]$. Consider the polynomial $f=xy-1$ in $R[x]$. Then the quotient $R/\langle f \rangle$ is isomorphic to $K[[y]][x]/\langle xy-1\rangle$, which is a field (it is isomorphic to the field of fractions of $K[[y]]$).
Thus the ideal $\langle f \rangle$ can be maximal.