Resistance in the context of electric loads
The electrical power put into a load, V*I, is transferred from the source of the voltage to the load. What happens to that energy put into the load depends, of course, on what that load is.

If the load is a simple resistor, then that energy just goes into producing heat. You are correct that for a pure voltage source with no internal resistance of its own connected to a simple resistor as a load, the higher the resistance, the less energy is dissipated in the resistor as heat (\$P=V^2/R\$). In this scenario, lowering the resistance just keeps increasing the power dissipated in the load.

The situation is more subtle when you consider that all real voltage sources have their own internal resistance. In that case, you find that you get the most energy into the load resistor if the resistance of the load equals the internal resistance of the voltage source (look up Maximum Power Transfer Theorem for more info). What happens is that the power is shared between the internal resistance of the source and the load resistance. If you lower the resistance of the load well below that of the source, then most of the energy from the source voltage just goes into the source resistance instead of the load resistance, and so lowering the load resistance no longer keeps increasing power dissipated in the load resistor.