Usually, lasers are not very efficient at converting input power to laser light. Most of it gets converted to heat, which needs to be dissipated somehow. If you made the reflectivity very high, then the laser would just stop outputting light, and instead would generate a little more heat. It probably wouldn't destroy the mirrors, since the laser wasn't very efficient to begin with, and increasing the heat a little probably won't overtax the heat sinks.
In a laser, you have a laser medium with multiple excited states. For simplicity, let's consider a three state system. G is the ground state. A is the upper state, and B is the middle state. The power supply is connected to some arc source which excites the laser medium to the A and B states. A decays to B faster than B decays to G, so you get a population inversion between B and the G. So, you get lasing for the B to G transition. But the A to B transition photons are basically lost somewhere, and I guess they get converted to heat eventually. If you turn up the reflectivity, you'll reach some steady state field energy in the cavity, and all the extra power gets converted to heat. Since there is no laser power being carried away, the laser medium becomes more excited and you have more A to B transitions creating heat.
