Prove that an infinite ring with finite quotient rings is an integral domain

If $ab=0$, and if $k_i$, $1\leq i\leq m$, are representatives mod $a$, then $(b)=\{bk_1,\dots,bk_m\}$, i.e. $(b)$ is finite and hence $R/(b)$ infinite.

I give a solution when $R$ is commutative and noetherian.

Suppose absurdly that $R$ is not a domain, hence $0$ is not a prime ideal. Therefore, for every prime ideal $\mathfrak{p}$ of $R$, the domain $R / \mathfrak{p}$ is finite, hence a field. So every prime ideal of $R$ is maximal, i.e. $R$ is artinian. Therefore $R$ is a direct product of some of its quotients, hence $R$ is finite.

Here's another way to look at it: suppose that $R$ is as you say, and assume by way of contradiction that $R$ is not a domain. Then $ab=0$ for some nonzero $a,b\in R$. Let $\operatorname{Ann}_R(b):=\{x\in R\colon xb=0\}$. Then both $R/Rb$ and $R/\operatorname{Ann}_R(b)$ are finite. Since $R$ is infinite, both $Rb$ and $\operatorname{Ann}_R(b)$ are infinite. But $Rb\cong R/\operatorname{Ann}_R(b)$, thus $Rb$ is finite, a contradiction.