Uncountable Reduced ring $R$ with $R[x]$ has only a countable number of maximal left ideals The question is following:

Is there an uncountable  reduced ring (i.e., a ring with no non-zero nilpotent element) $R$ (with identity) such that
  $R[x]$ has only a countable number of maximal left ideals?

 A: Let $r\in R$. By the Zorn lemma, there is a maximal left $R$-submodule $M_r$ in $R+Rx\cong R\oplus R$ such that
$M_r\left[\begin{matrix}0&1\\r&0\end{matrix}\right]\subset M_r$ and
$1\not\in M_r$. Denote $L_r:=R[x](x^2-r)+R[x]M_r$. As
$x^{2n}=(\sum_{i=0}^{n-1}r^{n-i-1}x^{2i})(x^2-r)+r^n$ and
$x^{2n+1}=(\sum_{i=0}^{n-1}r^{n-i-1}x^{2i+1})(x^2-r)+r^nx$, every element in $L_r$, being considered modulo $R[x](x^2-r)$, lies in $M_r$ and
$L_r=R[x](x^2-r)\oplus M_r$ as a left $R$-module. Moreover, $M_r=L_r\cap(R+Rx)$. Indeed, it suffices to observe that $M_rx=M_r\left[\begin{matrix}0&1\\r&0\end{matrix}\right]\mod R[x](x^2-r)$ and that $p(x)(x^2-r)\in M_r$ implies $p(x)=0$. Now it is easy to see that $L_r$ is a maximal left ideal in $R[x]$. Such ideals are different for many different $r$'s. Indeed, if $L_a=L_b$, then $a-b\in L_a=L_b$, hence,
$a-b\in M_a=M_b$. Consequently,
$\sum_{k\ge0}R(a-b)a^k+\sum_{k\ge0}R(a-b)a^kx\subset M_a=M_b$. We can take uncountably many choices of $a,b\in R$ and $M_a,M_b$ such that either
$M_a\ne M_b$ or
$\sum_{k\ge0}R(a-b)a^k+\sum_{k\ge0}R(a-b)a^kx\not\subset M_a=M_b$. [Here is a gap in the proof because of lack of time.] Therefore, there is no ring in question.
