The final thing that I can't evaluate is that if you double the voltage and keep the amperage the same, you will have doubled the wattage. When doing heat dissipation calculations, you calculate the wattage dissipated by the component. But when sizing wires based on heat build-up, you use amperage alone, regardless of the voltage (which allows getting more power through smaller wires by going to higher voltages). I don't know how to resolve this, though.
In typical power disapation calculations for sizing, either the voltage is fixed for the operation of the device (usually for something non-linear) or the impedence of the device is fixed (which in turn usually fixes the voltage/current tradeoff). Simply put, if I were to do this for a resistor (really any linear circuit), if I increase the current, the voltage must simultanously increase to force that current through. Current, voltage and power are inseperably linked.
This is not the case for an electric motor.
An electric motor will have a back EMF altering the total impedence depending on the load. *Assuming* the load were reactive to input voltage (the builder changes props between changes in voltage), such that current was kept constant, and *assuming* the voltage was kept well under the insulation's breakdown voltage, while the power transfered from the input source to beaten air would increase, the current would not . . . by definition.
An electric motor can have a variable current/voltage relationship, depending on the load.
As you've seen, it is the current flow that generates the waste heat, not the voltage -- a broken wire generates no electrical heat regardles of the voltage difference (up to the air's break-down voltage), while a near-0-volt hard short generates sufficent heat to melt and weld steel (and not only do the welding points where the current is concentrated heat up, so does the whole welding aparatus even though it's a very low resistance path).
While waste power can be used in calculating power transfer efficiencies -- it's convienient to express things in terms of power -- however do not expect the waste power to increase with voltage increases. there are emperical connections between current and heat, but voltage does not play in this.
Therefore, if you fix your current by varying your load with voltage, you'll find increasing the voltage will actually increase your power transfer efficiency because the waste component remains the same . . . because we've fixed the current.
To go back to the OP, and summarize (from Josh's posts) . . . so long as the prop size is changed to keep it efficient and limit current, the next failure mode is from overspeeding the bearings. If the bearings can take it, and the prop is sized right, kicking up the voltage will allow you to get proprtionally more thrust-power out of the same motor, at a greater overall efficiency.
I would not assume, however, that doubling the voltage will cut the required prop size in half. prop disk area, maybe-kinda-sorta, but size, no. Most of the Rules of Thumb I have at hand for comparing props are for a set voltage/RPM, and power draw for a given prop is very nonlinear with voltage . . . so I'd caution, test carefully with a power meter/current meter before you lock into that prop and fly extended periods WOT.