Suppose that f is nilpotent. Suppose P is prime ideal in A. Consider the ring $R = (A/P)[x]$. Consider the morphism given by
$$b_0 + b_1 + ... + b_nx^n \mapsto [b_0] + ... + [b_n]x^n.$$
Denote that morphism by $\phi$. As $\phi$ is a morphism it must map nilpotent elements to nilpotent elements. Therefore we must have $[a_0] + ... + [a_n]x^n$ must be nilpotent. Since $A/P$ is an integral domain so we must have $[a_i] = [0]$ for all i. So, $a_i \in P$ for all i. Thus $a_i \in Nil(R)$ as P was arbitrarily.
Conversely The set of nilpotent elements form an ideal. As $A \subset A[x]$, so if $a_i$ is nilpotent in $A \subset A[x]$. This means that $a_ix^i$ is nilpotent. Therefore $f = a_0 + ... + a_nx^n$ is nilpotent.