Nikola Tesla didn't believe in relativity. More historical context here. He made the following argument against general relativity in a 1931 interview with Hugo Gernsback:
Tesla contradicts a part of the relativity theory emphatically, holding that mass is unalterable; otherwise, energy could be produced from nothing, since the kinetic energy acquired in the fall of a body would be greater than that necessary to lift it at a small velocity.
Filling in some of the steps in the logic, it seems that he is assuming that the passive gravitational mass equals the relativistic mass, and therefore the force of the earth's gravity during the fall would be greater than the force during the slow lifting. That means the work around a closed path would be nonzero.
Is there a simple explanation for this? It's easy to pick holes in the argument, since GR doesn't describe gravity as a force, and gravitational interactions depend on the stress-energy tensor, not the mass-energy. But that doesn't feel like a complete resolution of the issue to me.
The relation $W=\int F dx$ is exact in special relativity if $F$ is the three-force, since $dE/dx=(dE/dp)(dp/dt)(dt/dx)$, and $dE/dp=p/E$. However, this seems ambiguous in the context of GR, since, e.g., the tension in a hanging piece of rope is subject to a correction equal to the gravitational redshift factor evaluated between the ends.
A possible way to get at this would be to imagine a slightly different scenario that may be simpler to reason about. We shoot a test particle straight down into a planet's gravity well with relativistic energy $E_1$. At the bottom, we reduce its speed to exactly escape velocity, extracting energy $E_2$ from it, and then reflect it back up, so that it rises back up with zero total energy. Then Tesla's argument would seem to be that $E_2>E_1$, which seems unlikely since we have a conserved energy for the geodesic motion of a test particle in this field.