Are there infinite sets of axioms? I'm reading Behnke's Fundamentals of mathematics:

If the number of axioms is finite, we can reduce the concept of a consequence to that of a tautology.

I got curious on this: Are there infinite sets of axioms? The only thing I could think about is the possible existence of unknown axioms and perhaps some belief that this number of axioms is infinite.

In first order Peano axioms the principal of mathematical induction is not one axiom, but a "template" called an axiom scheme. For every possible expression (or "predicate") with a free variable, $P(n)$, we have the axiom:
$$(P(0) \land \left(\forall n: P(n)\implies P(n+1)\right))\implies \\\forall n: P(n)$$
So, if $P(x)$ is the predicate, $x\cdot 0 = 1$ then we'd have the messy axiom:
$$(0\cdot 0=1 \land \left(\forall n: n\cdot 0 =1\implies (n+1)\cdot 0=1\right))\implies \\\forall n: n\cdot 0 = 1$$
Our inclination is to think of this axiom scheme as a single axiom when preceded by "$\forall P$", but in first-order theory, there is only one "type." In first-order number theory, that type is "natural number." So there is no room in the language for the concept of $\forall P$. In second order theory, we can say $\forall P$.
In set theory, you have a similar rule, the "axiom of specification" which lets you construct a set from any predicate, $P(x,y)$, with two free variables:
$$\forall S:\exists T: \forall x: (x\in T\iff (x\in S\land P(x,S)))$$
(The axiom lets you do more, but this is a simple case.)
which essentially means that there exists a set:
$$\{x\in S: P(x,S)\}$$
Again, there is no such object inside set theory as a "predicate."
For most human axiom systems, even when the axioms are infinite, we have a level of verifiability. We usually desire an ability to verify a proof using mechanistic means, and therefore, given any step in a proof, we desire the ability to verify the step in a finite amount of time.
