The spectrum of a product of operators

The answer to your question is no, the spectrum can be infinite, and pairs of self-adjoint involutions are actually a good class of counterexamples because they can be fully described using the spectral theorem. The irreducible pairs of involutions occur in dimensions $1$ and $2$, and the rest are direct integrals of irreducibles. So every such pair is described by the "spectrum of angles" between them with its spectral measure type (i.e. equivalence class) + multiplicity.

Now, an irreducible pair of involutions in dimension $2$ looks like this:

$A_\alpha = \left[\begin{matrix}1\\ & -1 \end{matrix}\right], B_\alpha = \left[\begin{matrix}\cos\alpha & \sin\alpha\\ \sin\alpha & -\cos\alpha \end{matrix}\right]$

The product $A_\alpha B_\alpha$ is the unitary operator that rotates by angle $\alpha$, so its spectrum is $\{e^{i\alpha}, e^{-i\alpha}\}$. Clearly, a direct sum of countably many of these, with different $\alpha$'s, can have infinite spectrum.

Incidentally, we see that every unitary operator whose spectral measure type and multiplicity function are symmetric under complex conjugation is a product of two self-adjoint involutions.