Is it true that in any game with 2 players each having only 2 pure strategies, there will be at most 3 Nash equilibria? Are there results on the bounds to the number of NE for general finite games?
The number of Nash equilibria of games

OK, obviously, I'm not interested in such cases.
Obviously? You asked a question, you got an answer to the question you asked.
I guess OP is looking for input on the following conjecture: in any 2by2 game, the possible numbers of mixed Nash equilibria are 1, 2, 3, or infinitely many. Generically, this number is 1 or 3. This actually strikes me as an interesting question.

I don’t know if there is a theorem that gives results for the bounds. I have a counterexample other than the zero payoffs case. I can’t think of a symmetric, non trivial 2x2 game in which there are more than 3 equilibria, though.
a1={T,B}
a2={L,R}The payoffs are
TL = (0,2)
TR = (0,2)
BL = (1,1)
BR = (1,1)Let us know what you find out, OP! I’m curious too.

I am a gametheorist but I never understood the point of results like that. What exactly do I get from learning that the number of equilibria is odd (a.s.)?
Also a game theorist. I think that the point of these such results is that it's an interesting property of a mathematical object.
Viewing game theory as a modeling tool, I agree, it doesn't seem terribly useful. Perhaps it is (or may be) useful in computing equilibria?

I am a gametheorist but I never understood the point of results like that. What exactly do I get from learning that the number of equilibria is odd (a.s.)?
When you find an even number of eqa either you found too many or too few. Or, you are in a very special caseand maybe it's worth to double check