Does having a finite number of generators, each having finite order, imply that the group is finite?

Consider the free group $G$ on $a$ and $b$, mod the relation that $a^2 = b^2 = e$. Then $G$ admits sequences of arbitrary length, namely $a$, $ab$, $aba$, $abab$, $ababa$, $\ldots$.

Define $f:\mathbb{Z}\to\mathbb{Z}$ by $$f(n)=\begin{cases} n+1&\text{ if $n$ is even} \\ n-1&\text{ if $n$ is odd}\end{cases}$$ and $g:\mathbb{Z}\to\mathbb{Z}$ by $$g(n)=\begin{cases} n-1&\text{ if $n$ is even} \\ n+1&\text{ if $n$ is odd.}\end{cases}$$

Then $f$ and $g$ are both permutations of $\mathbb{Z}$, and have order $2$ in as elements of the group of all permutations of $\mathbb{Z}$. But notice that $$(g\circ f)(n)=\begin{cases} n+2&\text{ if $n$ is even} \\ n-2&\text{ if $n$ is odd}\end{cases}$$ so $g\circ f$ is a permutation of infinite order (as you iterate it on even and odd integers, you just keep adding or subtracting $2$, respectively). It follows that the group of permutations generated by $f$ and $g$ is infinite, even though $f$ and $g$ both have finite order.

(In fact, it turns out that this is the same as silvascientist's example. The explicit representation as permutations makes it clear that all the words you can form by alternating $g$ and $f$ really are distinct elements, since you can just compute what they are as functions.)

In plane Euclidean geometry, the group generated two reflections in two parallel lines provides an easy example. The product of the two reflections is a nonzero translation, which has infinite order.

I. The dihedral group $D_n$ is a group of order $2n$ generated by two elements $a_n,b_n$ of order $2.$ Then $a=(a_1,a_2,a_3,\dots)$ and $b=(b_1,b_2,b_3,\dots)$ are elements of order $2$ in the direct product $\prod_{n=1}^\infty D_n,$ and they generate a subgroup $D=\langle a,b\rangle$ which is infinite because each $D_n$ is a homomorphic image of $D$ via the projection map.

II. Any permutation can be expressed as a product of two involutions [W. R. Scott, Group Theory, Prentice–Hall, 1964, Exercise 10.1.17]. In particular, given a permutation $\pi$ of infinite order, we can write $\pi=\alpha\beta$ for some involutions $\alpha$ and $\beta,$ and then $G=\langle\alpha,\beta\rangle$ is an infinite group generated by two elements of order $2.$ For a concrete example where $\pi$ is an infinite cycle in $\operatorname{Sym}(\mathbb N),$ take $$\alpha=(1\ 2)(3\ 4)(5\ 6)(7\ 8)\cdots,$$ $$\beta=(1)(2\ 3)(4\ 5)(6\ 7)\cdots,$$ $$\pi=\alpha\beta=(\cdot\ \cdot\ \cdot\ 7\ 5\ 3\ 1\ 2\ 4\ 6\ 8\ \cdot\ \cdot\ \cdot).$$

III. Any countable group can be embedded in a $2$-generator group with generators of prescribed orders $p\ge3$ and $q\ge2$ [F. Levin, Factor groups of the modular group, J. London Math. Soc. 43 (1968), 195–203]. Hence, any countable group can be embedded in a group generated by three elements of order $2.$

IV. See Burnside's problem for examples of finitely generated infinite groups in which each element has finite order.