If $R$ is an integral domain, then $(R[x])^\times=R^\times$ If $R$ is an integral domain, then $(R[x])^\times=R^\times$

So since $R$ is an integral domain, it follows that $R[x]$ is an integral domain. We have $f(x)g(x)=1$ then we know that $\deg(f(x)g(x))= \deg(f(x))+\deg(g(x))=0$
Since neither can be equal to zero, they are both non-zero degree $0$ polynomials, and therefore $f(x)=a,g(x)=b$ where $f,g\in R$. Hence all units in $R[x]$ must be units in $R$.

Is this acceptable?

Your proof seems fine.
Using your result try to prove the following general result:

Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.

Try to prove this result on your own. (Hint. Assume the result for domains and for any prime ideal $p$ of $R$ and for any polynomial $f(x)$, consider $ \bar f(x)$ in the integral domain $(R/p)[x]$ by reducing coefficients modulo $p$ and use the result for integral domains.)
