The Helmholtz equation is  ∆u + ßu = f, so after discretization you get  Au + ßu = f, where A is a discrete (say finite difference) approximation to the Laplacian. Clearly, the eigenvalues of (A + ß)  =  ß + eigenvalues of A. So it all boils down to finding the eigenvalues of the discrete Laplacian (+ boundary conditions). Depending on the computational domain and boundary conditions you can find the discrete eigenvalues of the Laplacian in the literature. For example, have a look at the classroom notes

http://web.stanford.edu/class/math220b/handouts/eigenvalues.pdf

Are you asking for lower bounds or upper bounds?

I don't think that there are upper bounds for the eigenvalue problem. In the discrete version you might have one because the number of eigenvalues is finite. Now, because of the variational nature of your problem, the approximated eigenvalues might be greater than the exact eigenvalues... And you can think about them as an upper bound for the exact case.

I don't know of any lower bounds, although that would be interesting to have.

You can see the eigenvalues spectra of the Helmholtz operator in my recent work. I have solved the Helmholtz equation with null Dirichlet BC using an improved global meshless method (RBF-PS). You can see the eigenvalues for various degrees of freedom.

Although this eprint is a first draft, I think it might be useful in this context.

http://arxiv.org/abs/1606.03258