Question

show the parity operator can have only two eigen values

Answer 1

You do not need to invoke a theorem in order to prove that if an operator commutes with the identity operator, then they have common eigenvectors. That's trivial, since any (non-zero) vector is an eigenvector of the identity operator and any operator commutes with the identity operator.

So, the answer to your question is “yes”, but it's quite trivial. Indeed, there are infinitely many eigenvectors of the parity operator and each of them is an eigenvector of the identity operator. So, both operators have infinitely many eigenvectors.