Zev's argument in fact shows that there cannot be any finite subrings in a ring of characteristic zero (when subrings are required to have the same multiplicative unit), and Niel's answer shows that there can be a non-trivial finite subrng in an infinite ring.
So, for an example of a finite subring of an infinite ring, let us look at rings of positive characteristic. For example, let $\mathbb{F}_p$ be the finite field of $p$ elements. Then it is a subring of the polynomial ring $\mathbb{F}_p(X)$, which is infinite. It is also a subring of the algebraic completion $\overline{\mathbb{F}_p}$, which is an infinite field.
There are also infinite rings which contain no non-trivial finite subrngs: $\mathbb{Z}$ is an example, since every additive subgroup is infinite. In fact, every integral domain of non-zero characteristic cannot contain any non-trivial finite subrngs. Let $R$ be such a ring, and suppose $S$ is a non-trivial additive subgroup. Then, $S$ contains a non-zero element $x$. But $R$ contains an isomorphic copy of $\mathbb{Z}$, and $R$ is an integral domain, so for any non-zero integer $n$, $n x \ne 0$. In particular, if $n \ne m$, then $n x \ne m x$, and both are in $S$ if $n$ and $m$ are integers. So $S$ is infinite.