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.