astrophysics - Does conservation of energy make black holes impossible?

For simplicity, let's consider a Schwarzschild black hole, so that the spacetime is spherically symmetric and static. In particular, the Schwarzschild time $t$ coordinate gives a direction in which the geometry 'stays the same' (a Killing vector field), and its inner product with the orbital four-velocity $u$ is conserved:
$$epsilon = -g(u,partial_t) = left(1-frac{2M}{r}right)frac{mathrm{d}t}{mathrm{d}tau}text{,}$$
where $tau$ is the proper time of the orbiting particle. One can think of this as the specific (per-mass) energy, including mass-energy: a particle escaping to infinity that becomes asymptotically at rest with respect to stationary observers would have $epsilon = 1$.

If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite.

The energy as measured by whom? Imagine a family of stationary observers everywhere surrounding the black hole, or at least along an infalling particle's trajectory. Those observers measure the speed and energy of the particle as it falls past them. As the particle nears the horizon, they will report speeds arbitrarily close to the speed of light and arbitrarily high energies.

But to a far-away stationary observer, those observers near the horizon are experiencing increasingly divergent gravitational time dilation, and the orbital energy of the particle stays constant. Flinging a particle of mass $m$ into the black hole will increase the mass of the black hole by $mepsilon$.

What's the critical flaw in this argument?

All energy gravitates, so I would say the main flaw is forgetting to add the gravitational potential energy of the particle to the mass of the black hole as well. Or better put, you should be concerned with the orbital energy, not just its mass+kinetic parts ('relativistic mass').

In this situation, you have a well-defined conserved orbital energy. If you insist on measuring the Lorentz factor according to close-by stationary observers, then yes, it diverges to $+infty$, but then you would have to admit a gravitational potential energy term that diverges to $-infty$, because their sum must resolve to be the orbital energy.