I shall assume all rings to be commutative in this question. The impatient can scroll down to the "blockquote" to read the actual question.
Whenever we have a polynomial over a ring, it defines a function from the ring to itself by evaluation. It's reasonable to ask when two different polynomials define the same function.
From the factor theorem it follows that an $n^\text{th}$ degree polynomial over an integral domain has at most $n$ roots. Then it's easy to show this:
Theorem. Let $R$ be an infinite integral domain and let $f \in R[X]$ such that $f(a)=0$ for all $a \in R$, then $f = 0$.
Proof. $f$ has infinitely many roots, so it must be the zero polynomial. $\quad\square$
For finite rings a kind of opposite situation occurs:
Theorem. For any finite ring $R$ there are polynomials over $R$ that are different but agree on all elements.
Proof. There are only finitely many functions from $R$ to itself, but $R[X]$ is infinite. $\quad\square$
If we make further assumptions it's of course possible to prove more, as Pete L. Clark wrote in this post: [1]
Then there is the question of infinite rings that are not integral domains. It's relatively easy to come up with examples of a ring $R$ with positive characteristic and a nonzero polynomial that evaluates to the zero function, e.g.: $ R := \bigoplus_{n=1}^\infty \mathbb{Z}/6\mathbb{Z} \quad\text{and}\quad f(X) := X^3-X.$
The Question:
This leaves open the case alluded to in this post's title: Is there a commutative ring of characteristic $0$ (hence infinite) such that a nonzero polynomial evaluates to the zero function?