You are correct that if you can prove that $R[x]/\langle x\rangle$ is a field, then you can conclude that $\langle x\rangle$ is maximal.
However, it is not true in general that if $a$ is irreducible in a domain $D$, then $D/\langle a\rangle$ is a field. For example, $2$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$, but $\langle 2\rangle$ is not a maximal ideal of $\mathbb{Z}[\sqrt{-5}]$. So you need more than just "$f(x) = x$ is irreducible in $R[x]$" to conclude "$R[x]/\langle x\rangle$ is a field."
(For another, simpler example, consider the case of $\langle x \rangle$ in $R[x,y]$. You can still say that "since $R$ is a field, $x$ is irreducible", but it is no longer true that $R[x,y]/\langle x\rangle$ is a field; in fact, $\langle x\rangle$ is not maximal in this ring, since it is properly contained in the proper ideal $\langle x,y\rangle$.)
But perhaps you can show that the homomorphism $R[x]\to R$ given by "evaluation at $0$" is onto, and has kernel $\langle x\rangle$? That will show that $R[x]/\langle x\rangle \cong R$ is a field.