Let us assume that once the bubble is created, the physics is essentially no different from what is responsible for the work making an Helium balloon rise in the air i.e. you have a macroscopic/mesoscopic object which has a density smaller than that of the fluid surrounding it.
Now, we all know that in a static fluid under gravity, pressure is decreasing with height. The reason for that is because any fluid follows an equation of state $p = f(n,T)$ where, at fixed temperature say $f$ is an increasing function of the particle density $n$.
Hence saying that the pressure decreases as we go higher in a fluid means that the density goes lower.
If the temperature is the same (buoyancy is observed on length scales much smaller than the scale over which temperature changes) then, it means that the big particle doesn't receive the same amount of collisions coming from the bottom than that coming from the top. As a consequence it will experience a net force in the upward direction.
As Pratyay Gosh has said, the energy of the system need not change here, the thing that changes however is the total entropy of the system that is related to the fact that closer to the bottom wall there are more particles that are "bothered" by the presence of the lighter bubble (because it's huge) than if you were to put this bubble at a higher position.
One way to realize that is to imagine the amount of work one needs to perform to insert a bubble somewhere at some height in the fluid. It is kind of intuitive to see that the less dense the fluid is, the more probable it is to find a "void" in which one can insert the bubble. That is what I mean when I say that is because of entropy.