This is question 7.4.6 from Dummit and Foote. Just need someone to verify my answer.
$$R[x]/(x) \stackrel{\pi}\leftrightsquigarrow R$$
$$a_0 + (a_1 + \dots a_nx^{n-1})x = a_0 + (x) \stackrel{\pi}\leftrightsquigarrow a_0$$
This is clearly an isomorphism with kernel $\ker \pi = (x) = \bar{0} \in R[x]/(x)$. This will answer both question since if $(x)$ is maximal, it is prime. Then $R[x]/(x)$ is a field and hence $R$. Similarly $R[x]/(x)$ would be an integral domain, and so is $R$.
Well, I'd consider the substitution homomorphism $\phi:R[x]\rightarrow R$ defined by $\phi:f(x)\mapsto f(0)$. The kernel is the ideal $I=\langle x\rangle$ and so by the homomorphism theorem, $R[x]/\ker(\phi) = R[x]/\langle x\rangle$ is isomorphic to $\phi(R[x])$ which is $R$, since $\phi$ is surjective. The isomorphism is defined by $f(x)+\langle x\rangle \mapsto f(0)$.