Identifying units in a polynomial ring 
Problem Statement: Let $R$ be a domain. Identify the units in $R[x]$.

I am trying to identify the units in a domain $R$ by considering an arbitrary element $a=a_{n}x^{n}+\cdots+a_{1}x+a_{0}\in R[x]$ with an inverse $b=b_{m}x^{m}+\cdots+b_{1}x+b_{0}\in R[x]$.
If $b=a^{-1}$, then $$1=ab=ba=(a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0})(b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0})
 = (a_{n}b_{m}x^{n+m}+a_{n}b_{m-1}x^{n+m-1}+\cdots+a_{n}b_{1}x^{n+1}+a_{n}b_{0}x^{n})
+ (a_{n-1}b_{m}x^{n+m-1}+a_{n-1}b_{m-1}x^{n+m-2}+\cdots+a_{n-1}b_{1}x^{n}+a_{n-1}b_{0}x^{n-1})+\cdots
+ (a_{1}b_{m}x^{m+1}+a_{1}b_{m-1}x^{m}+\cdots+a_{1}b_{1}x^{2}+a_{1}b_{0}x)
+(a_{0}b_{m}x^{m}+a_{0}b_{m-1}x^{m-1}+\cdots+a_{0}b_{1}x+a_{0}b_{0})$$
$$= a_{n}b_{m}x^{m+n}+(a_{n}b_{m-1}+a_{n-1}b_{m})x^{n+m-1} +
 (a_{n}b_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_{m})x^{n+m-2}+\cdots+(a_{2}b_{0}+a_{1}b_{1}+a_{0}b_{2})x^{2}+(a_{1}b_{0}+a_{0}b_{1})x+a_0b_0$$
Thus,
$$a_{n}b_{m}=a_{n}b_{m-1}+a_{n-1}b_{m}=a_{n}b_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_{m}=\cdots=a_{2}b_{0}+a_{1}b_{1}+a_{0}b_{2}=a_{1}b_{0}+a_{0}b_{1}=0$$
Or more generally,
$$a_{i}b_{0}+a_{i-1}b_{1}+\cdots+a_{1}b_{i-1}+a_{0}b_{i}=0$$
and
$$a_{0}b_{0}=1.$$
Since $R$ is a domain, $R[x]$ is a domain, and thus, $a_{n}b_{m}=0$ implies that $a_{n}=0$ or $b_{m} =0$.
If $a_{n}=0$ and $b_{m}\neq0$, then it would recursively follow that $a_{n}=0\ \Rightarrow a_{n-1}=0$?
Similarly, If $b_{m}=0$ and $a_{n}\neq0$, then it would recursively follow that $b_{m}=0\ \Rightarrow b_{m-1}=0$?
But that would imply that either $a_{0}=0$ or $b_{0} = 0$, which is an issue, because we want $a_{0}b_{0}=1$.
I am thinking that the only units are constant polynomials? That is, unless there is some positive integer $k$ such that $x^{k}=0$. Should I be considering that case as well, or is there an easier way to approach this problem?
Any suggestions are appreciated!

Hint: if $R$ is a domain, then $\deg{fg} = \deg{f} + \deg{g}$ for all polynomials $f, g\in R[x]$.
