Maximal ideals in $\mathbb{R}[X]$

Irreducibility. $X^2+1$ has no real root (discriminant $-4<0$), so it is irreducible over $\mathbb{R}$.

Maximality. Let $I$ be an ideal with $(P)\subseteq I\subseteq\mathbb{R}[X]$. Since $\mathbb{R}[X]$ is a PID, $I=(Q)$ for some $Q$. Then $P\in(Q)$, so $Q\mid P$. Since $P$ is irreducible, $Q$ is a unit (then $I=\mathbb{R}[X]$) or $Q$ is an associate of $P$ (then $I=(P)$). So $(P)$ is maximal.

Conclusion. $(X^2+1)$ is maximal. Moreover, $\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}$ is a field.