When using such equations we must be careful which voltage, which current and which resistance we are talking about.
Lets assume a simple case, the supply is DC so there are no capacitive or inductive affects and the cable has perfect insulation but the conductors have some resistance. Now lets write some equations.
$$V_{load}=V_{source}-V_{drop}$$
Where \$V_{load}\$ is the voltage delivered to the load V_{source}\$V_{source}\$ is the voltage supplied by the source and V_{drop}\$V_{drop}\$ is the voltage dropped in the cable.
simulate this circuit – Schematic created using CircuitLab
By ohms law we can write
$$V_{drop} = I * R_{cable}$$
Where \$I\$ is the current flow in the circuit and\$R_{cable}\$ is the total resistance of the conductors (both positive and negative) in the cable supplying the load. We can now write an equation for the power loss in the cable.
$$P_{loss} = V_{drop} * I = I^2 * R_{cable} = V_{drop}^2 / R_{cable}$$
So to reduce \$P_{loss}\$ we need to either reduce \$R_{cable}\$ or reduce \$I\$. To reduce \$I\$ while keeping the power delivered to the load the same we have to increase \$V_{load}\$ and hence \$V_{supply}\$
Now this example isn't a perfect reflection of the real world. In reality insulators aren't perfect and systems are usually AC so capacitive and inductive affects have to be considered. The result is that increasing voltage helps up to a point but eventualy you reach a point where further voltage increases are not helpful.
