I (Anton) have edited this question to be the question Pete and Zeb discuss in the first few comments.
What conditions on a ring $R$ imply that the units of $R[x]$ are exactly the units of $R$?
I (Anton) have edited this question to be the question Pete and Zeb discuss in the first few comments.



If $R$ is a commutative ring, then by the following result, the answer is "if and only if $R$ is reduced."
Proof. One direction is easy. Any polynomial of the given form is a unit because the sum of a unit and a nilpotent element is always a unit. The other direction isn't too hard if $R$ is a domain (the product of nonzero elements is always nonzero). If $g=b_0+\cdots b_mx^m$ (with $b_m\neq 0$) is the inverse of $f=a_0+\cdots+a_nx^n$ (with $a_n\neq 0$), then the highest order term of $1=f\cdot g$ is $a_nb_mx^{n+m}$, so we must have $n=m=0$ and $a_0$ invertible (with inverse $b_0$) For the general case, suppose $a_0+\cdots +a_nx^n$ is a unit. Reducing modulo $x$, we must get a unit in $R[x]/(x)\cong R$, so $a_0$ must be a unit. Reducing modulo any prime $\mathfrak p\subseteq R$, we get a unit in $(R/\mathfrak p)[x]$. Since $R/\mathfrak p$ is a domain, the previous paragraph shows that $a_i\in \mathfrak p$ for all $i>0$ and all primes $\mathfrak p$. Since the intersection of all primes is the nilradical, each $a_i$ must be nilpotent. A more "bare hands" elementary proof is given in Ex. 1.32 of Lam's Exercises in Classical Ring Theory. He also gives counterexamples to both implications if $R$ is not assumed commutative and mentions a really interesting related question. If $I\subseteq R$ is an ideal all of whose elements are nilpotent and $a_i\in I$, then does it follow that $1+a_1x+\cdots +a_nx^n$ is a unit in $R[x]$? If you can prove that it does, it would imply the Köthe conjecture, a famous problem in ring theory. 

