I have a game where there is NO equilibrium in pure strategies and there are two and ONLY TWO candidates for the mixed strategy Bayesian-Nash equilibrium. I can prove that one of them is indeed a mixed strategy equilibrium. However, I cannot disprove that the other one is not (I tried to find this equilibrium numerically for a wide range of parameter values, but no success).
I know that there must be only odd number of Nash equilibria. Can I claim that (i) I have only two candidates for equilibrium, (ii) I prove that one of them is indeed an equilibrium, (iii) since the number of equibria must be odd, the second candidate cannot be an equilibrium?
If these candidate equilibria are totally mixed, your argument holds, and without any recourse to even/odd stuff. In particular, to extend mlc's remark to all finite strategic-form games, rather than 'just' generic ones, it can be shown that the set of equilibria on the interior of strategy space of the game (the product of the individual agents' strategy simplexes) is always connected (intuitively, this should make sense in light of von-Neumann Morganstern utilities being affine and all).$^{1}$
Hence if you know that:
The set of equilibria lies within your two-point set; and
Your two-point set is contained in the interior of the strategy space,
then there is a unique equilibrium without further ado.
$^{1}$ https://mathoverflow.net/questions/89036/why-are-nash-equilibria-inside-the-simplex-s-n-unique