If one player always goes all-in or folds, then they are forcing the other player to almost do the same. So I wonder if there's a GTO strategy for HU NL hold'em in which the player only either folds or shoves, or if every such strategy is somehow exploitable. I'm also curious about the more general case of having more than 2 players.

I hope this will answer your question.

I think you're talking about Nash equilibrium for push/fold decisions heads-up. There is plenty of Nash charts on the internet

In a few words, if you play according to Nash equilibrium, best way your opponent can counteract your moves is playing Nash equilibrium himself.

Of course Nash equilibrium can be extended to a variety of different scenarios and number of players

No, i'm asking whether a Nash equillibrium strategy for NL holdem needs to use call, check or raise without shoving. There are Nash charts for push/fold decisions heads-up yes, so makes me wonder: if you always only either push / fold, you are forcing your oponent to either call (allin) or fold, so shouldn't there be such a strategy that is optimal?

I also asked this on poker.stackexchange and if the answer is correct, this is wrong because the other player can raise on the button: https://poker.stackexchange.com/a/11191/7829

If you're relatively short stacked (lets say 8bb or less) it starts making a lot more sense to only shove or fold.... At higher stack depths it's higher EV to have a limp + raise range rather than solely jamming.

The stackexchange answer seems pretty good, so I'm not sure exactly where your confusion is.

Push/fold Nash Equilibria are not full game equilibria. Push/folding forces you to play hands suboptimally which would have preferred to raise with a smaller sizing or limp. For example 44 is probably not a profitable shove 100BB deep, so you'd be forced to fold that hand for 0 EV when you could have raised it to a normal amount for +EV.

Edit: On stackexchange you're suggesting that a strategy which is >= 0EV against any other strategy would be optimal. That's not true. An optimal strategy maximizes EV against its best counterstrategy. You want the most +EV strategy, not just any strategy that wins money.

Suppose a strategy S is >=0 EV against any other strategy. Then in particular it is =0 EV against it self. So wouldn't the best counterstrategy against S be S itself, and therefore both definitions of optimal strategy equivalent?

The highest EV strategy against a strategy not S, does not equal S, if S is >= any strategy.

I see, i'm not very familiar with game theory terms, when i said optimal i meant it to mean EV>=0 against any other strategy, is there a term for that? Because in general against an arbitrary strategy we can't find out what the opponent's strategy is (i'm also not sure if my definition of strategy matches the game theoretic one: i'm assuming that it would be a valid strategy for, the first time a situation occurs to have a certain distribution of actions and the next time have a different one. by situation i mean position, stacks, hole cards, public cards, oponents previous decisions, etc)

I think they could beat you by playing your strategy in the BB, but min-raising or completing a lot in the SB.

Against a player not playing GTO, the maximally exploitative strategy (I.E. highest EV) is not GTO

Using a Nash strategy is only optimal if you are only allowed to go all-in or fold.

A simple extreme example to illustrate is if you have an effective stack of 1000 bb's. In that case you will be folding every hand that is not AA so your opponent can easily beat your strategy by grinding you down.

As mentioned, with 8 bb's or less using this strategy is a lot closer to a GTO solution but still doesn't mean that there isn't a better solution by limping/minraising some hands