Let $R\ne(0)$ be a commutative unital local ring (it has a unique maximal ideal that we'll call $\mathbb{m}$). In order to show that $x\notin\mathbb{m}\implies$ $x$ is a unit I first want to prove that for a ring (not necessarily local) [$x\in$ a maximal ideal] $\iff$ [x is not a unit].
$(x)$ is the ideal generated by $x$ in $R[x]$
Here is how I did one implication:
$x$ is a unit $\implies$ $\exists y\ :\ xy=yx=1\implies 1\in(x)$ holds $\implies$ $R\subset (x)=\{a_0+a_1x+...+a_nx^n\ :\ n\in\mathbb{N}\}$ so if an ideal contains $x$ it has to contain $1$ and so it cannot be maximal
with this result it is easy to prove that $x\notin\mathbb{m}\implies$ $x$ is a unit:
$x\in R\backslash\mathbb{m}$ and let's assume that there is no $y\in R$ s.t. $xy=1$. By the previous result (the implication that I didn't do) $x$ is in a maximal ideal, but $\mathbb{m}$ is the unique maximal ideal which is absurd.
Could someone point out if there is something wrong with my reasonning, and give a hint for the other implication please?
And I want to stress that although there might be a far superior proof, I would really appreciate it if I could get some feedback on what I did.
You are complicating what is simple. After proving that $1\in(x)$, all you have to do in order to prove that $(x)\supset R$ is to note that, if $r\in R$, then $r=r\times1\in(x)$, since $(x)$ is an ideal.
The rest looks fine to me.
In order to prove the other implication, you can do this: let $x$ be a non-unit. Then $(x)$ is an ideal, and therefore it is contained in a maximal ideal. But here the only maximal ideal is $\mathrm m$. THerefore, $(x)\subset\mathrm m$. In other words, $x\in\mathrm m$.