Prove that $f = a_nx^n + \cdots + a_1x+a_0$ is a unit in $R[x]$ if and only if $a_0$ is a unit in $R$ and $a_1, ..., a_n$ are nilpotent.

My argument is as follows.

Suppose that $f$ is a unit in $R[x]$ and consider the evaluation map $\varphi : R[x] \to R$ defined by $x \mapsto 1$. We therefore have that $\varphi(f) = (a_n + \cdots + a_1) + a_0$. We note that Question 1 gives us that the sum of a nilpotent element and a unit is a unit. We therefore conclude that since $\varphi$ must map a unit in $R[x]$ to a unit in $R$, $(a_n + \cdots + a_1) + a_0$ is a unit in $R$. Furthermore, with the some of nilpotent elements being nilpotent, this yields that $a_1, ..., a_n$ are nilpotent and $a_0$ is a unit in $R$.

My argument lacks rigour in showing that $(a_n + \cdots a_1)$ must be nilpotent and $a_0$ must be the unit. I simply state this, which I'm not satisfied with. Could I take another evaluation map, this time evaluating at $x=0$?

Furthermore, the other direction seems unclear given that I cannot consider something as simple as an evaluation map.

Also note that this question is not a duplicate. The induction argument that is used almost ubiquitously as a solution is not intuitive and therefore I want to prove it through this approach of evaluation maps.