Is ZFC without Axiom of Infinity consistent?

The corollary from the incompleteness theorems is that you cannot prove the consistency of $\sf ZFC$ from $\sf ZFC$ itself. You have to have a stronger theory.

For example, in $\sf ZFC+\text{There exists an inaccessible cardinal}$, you can in fact prove the consistency of $\sf ZFC$ because this is a stronger theory.

Similarly this is the case of $\sf ZF_{fin}$ ($\sf ZF$ without infinity). The theory itself cannot prove its own consistency. However $\sf ZFC$ is a strictly stronger theory, and it proves the consistency of $\sf ZF_{fin}$. It does so by exhibiting a set which is a model of the theory, $V_\omega$ - the set of the hereditarily finite sets.

Large cardinal axioms are often called "strong infinity axioms" because they mimic the axiom of infinity, in the sense that they make a stronger theory by describing that a certain set of ordinals exists.

ZF without the Axiom of Infinity is still strong enough to derive Peano's axioms PA. So we cannot prove its consistency without proving the consistency of PA.