In THIS PROBLEM it is proved that for any nonzero ring $R$ with identity, $R[x]$ has an infinite number of maximal left ideals. Is it possible for an uncountable nonzero ring $R$ with identity, $R[x]$ has only a countable number of maximal left ideals ??
Generalizing the answer by wccanard: Let $R$ be a commutative ring. Then the kernel of $R[x] \twoheadrightarrow R_{red}[x]$ consists of nilpotent elements, hence this map induces a homeomorphism $\mathrm{Spec}(R_{red}[x]) \cong \mathrm{Spec}(R[x])$. It restricts to a homeomorphism between the subspaces of closed points $\mathrm{Spm}(R_{red}[x]) \cong \mathrm{Spm}(R[x])$. Now take any uncountable $R$ such that $R_{red}$ is a finite field. 


Yes. Take your favorite finite field, now form $R$ by adjoining uncountably many new variables each with square equal to zero and product of any two equal to zero (more precisely, let's say any element of $R$ is only allowed to mention finitely many of these variables) and now any maximal ideal will have to contain all of them, so done. 

