Higher voltage means lower current for a given power requirement. Lower current means less power loss to heat. The power required at the propeller is what is left over after the power lost in transmission is removed. Another way of saying this is that the power the battery must supply is what the propeller requires plus the power lost in transmission. In an ideal world where we could make batteries exactly to fit the application, which we can't, the higher voltage battery suitable for the mission profile will always be lighter.. The trouble we find is that all manufacturers of battery pack play games with C rating and MAH derating certain pack for greater C rating visa versa. What we truly need to know and understand is the total energy capacity of the pack or cell. in general the 18650 cells have a greater energy density than the poly cells. Trouble with them is they are often bigger than necessary for a specific mission. For the mathematically inclined:

      Power supplied by the battery = power required by the propeller + the power lost to heat

      Power supplied by the battery = (Torque x RPM/ 5252) + (Current x Total resistance (Wires, motor, back EMF...) )

      Voltage x MAH = ((Torque x RPM/ 5252) + (Current x Total resistance)) x Time

Note, the MAH on the pack doesn't mean a whole lot. Some manufacturers put the Watt Hour rating on the pack. That's the value we need to compare. Battery A Whr = Battery B Whr. You can get that roughly by the product of the pack voltage time the MAH rating. However it's the chemistry that makes the big difference. The 18650 cells win here as well as they also have the best energy per volume.

Here's how this impacts our model application. The parasitic power, the energy lost to heat that does us no good, is a function of the square of the current. Simply put is is very hurtful to our cause because of the square term. Write the first equation:

     Pbattery = Pprop + Plost

Power in an electrical system:

     Power = Current x Voltage,   where Voltage = Current x Resistance

Then:

     Plost = I x V =  I x (I x Rttl ) = I^2 x Rttl

Now to understand how this impacts us and drives us to always want to use the highest voltage for a specific size and weight we solve for the current:

     I = Square root ( Plost / V )

This is kind of a circular definition but you can see that the current is vary much driven inversely to the voltage. The higher the voltage to lower the current by a very significant amount. Keep in mind this current thing is what drives the heat up and down. Batteries get hotter, wires get hotter, MOTORS get hotter causing issues with plastic mounts....

So, short version. Always use the highest voltage you can. To compare two packs, multiply their voltage by their MAH. IF you know what one configuration is that works, use that V x MAH and then use the comparing packs V x MAH. If the latter is very mush greater than the former, look to see if there is smaller cell pack.

My $0.02 worth what paid for.