An observable only needs to be Hermitian, and can have any real eigenvalues. They don't even need to be distinct eigenvalues: if there are repeated eigenvalues, we say that the eigenspace for that eigenvalue is degenerate.
(In the case of observables on a qubit, having a repeated eigenvalue makes the observable rather uninteresting, because absolutely all pure states are eigenstates in that case; I'd be tempted to call such an observable 'degenerate' in an informal sense as well in that case — though it is on occasion useful to include $\mathbf 1$ in an analysis of things to do with single-qubit observables.)
In the analysis of Bell's Theorem, the reason why the observables are taken to be ones with eigenvalues $\pm1$ are conventional. It makes them analogous to Pauli spin operators in particular, and it makes a perfectly mixed state have expectation value $0$. Having eigenvalues of $\pm1$ also allows the expectation values of the operators to describe a bias towards one of two outcomes, and for expectation values of tensor products to be straightforwardly interpreted as a correlation coefficient of outcomes. You could prove versions of Bell's Theorem for observables with other eigenvalues, but those versions could be derived from Bell's Theorem as its usually stated.