Proposition. Let $R$ a commutative ring with identity. Let $M\ne R$ ad ideal, then $$\bigg (M\;\text{is maximal}\bigg)\iff \bigg(\forall r\in R\setminus M, \exists a\in R\;|\;1+ra\in M\bigg)$$
Proof. $(\Rightarrow)$ Let $r\in R\setminus M$ since $M$ is maximal $R=(M,r)=M+(r)$. Then $-1\in (M,r)$, therefore $\exists m\in M$ and $a\in R$ such that $-1=m+ra$, then $-m=1+ra\in M$.
$(\Leftarrow)$ Let $r\in R\setminus M$ we observe that $M\subset (M,r)\subseteq R.$ We note that $1+ra\in R$, but for hypotesys $1+ra\in M\subset (M,r)$, then $R\subseteq (M,r)$. Therefore $R=(M,r)$, and then $M$ is maximal.$\quad\square$
Question 1. Is the proof of this proposition correct?
Theorem. Let $R$ be a commutative ring with identity. There is a one-to-one correspondence between the maximal ideals $M$ of the ring $R$ and the maximal ideals $M'$ of $R[[x]]$ in such way that $M'$ corresponds to $M$ if and only if $M'$ is generated by $M$ and $x$; that is, $M'=(M,x).$
Proof. We write explicitly $M'=(M,x)$. $$M'=(M,x)=M+(x)=\{m+(\tilde{a_0}+\tilde{a_1}x+\cdots+)x\;|\;m\in M\}.$$
Let $M$ an maximal ideal in $R$. To prove that $M'=(M,x)$ is a maximal ideal of $R[[x]]$ is sufficient to prove that considered a formal series $f(x)=\sum_ka_kx^k\notin M'$, then $1+f(x)g(x)\in M'$ for some $g(x)\in R[[x]].$ Since $f(x)\notin M'$, the costant term $a_0\notin M$, in conseguence of this being $M$ maximal for the above proposition $\exists r\in R$ such that $1+ra_0\in M.$ Therefore \begin{equation} \begin{split} 1+rf(x)&=1+r(a_0+a_1x+a_2x^2+\cdots)x\\ &=(1+ra_0)+r(a_1+a_2x+\cdots)x\in (M,x). \end{split} \end{equation} So $M'$ is maximal.
Now, take $M'$ to be any maximal ideal of $R[[x]]$ and define the set $M$ to consist of the constant terms of power serie in $M'$: $$M=\bigg\{a_0\in R\;|\;\sum_k a_kx^k\in M'\bigg\}.$$
Question 2. How can I show that M is a maximal ideal of $R$?
We observe that $M\ne R$, in fact if it were $M=R$, then $1\in M$, therefore exist a power series $t(x)=\sum_k b_kx^k$ in $M'$ such that $1=1+0x+0x^2=b_0=b_1x+\cdots+$, then $b_0=1$. Therefore $b_0$ is invertible then $t(x)$ is invertible (this is a preliminary result that I had prove.) In conseguence of this $M'=R[[x]]$ which is impossible, because $M'$ is maximal.
Then it shows that the corrispondence is one to one.
Thanks for the patience!
The first proof is correct.
For the second part, you can observe that $$ R[[x]]/(M,x)\cong R/M $$ by considering the obvious homomorphism $R\to R[[x]]/(M,x)$, $r\mapsto r+(M,x)$. Prove that it is surjective and its kernel is $M$.
Thus if $M$ is maximal, then $(M,x)$ is maximal.
Conversely, suppose $I$ is a maximal ideal in $R[[x]]$. Consider $M=\{f(0):f\in I\}$ (that is, the set of all constant terms of the elements of $I$) and prove that $M$ is maximal in $R$. By the theorem, it is sufficient to prove that if $r\in R\setminus M$, there exists $a\in R$ with $1+ra\in M$.
Take $r\in R\setminus M$. Then $r$ as an element of $R[[x]]$ doesn't belong to $I$, so there exists $f\in R[[x]]$ with $1+rf\in I$; taking the constant term, we see that $1+rf(0)\in M$.
Now it's sufficient to prove that $I=(M,x)$. Take $f\in I$; then $f(0)\in M$ and $f=f(0)+x(f-f(0))\in (M,x)$; thus $I\subseteq(M,x)$. Since $(M,x)$ is maximal as well as $I$, we have $I=(M,x)$.