I know how to solve this using P=V²/R, but I don't understand why we can't use P=VI for the same. Why can't we?
TL;DR - there is an unstated assumption in the exercise that the resistance of the loads is constant.
For a device, the power stated is given at its nominal voltage, so P=VI would give you the current at the full 120V. For passive loads, if the supply voltage drops, the current goes down and the power goes down more.
For the heater and bulb, the exercise probably expects you to assume the resistance is constant, because they don't tell you anything about them. In reality, the operating resistance of the bulb and heater at a lower voltage than nominal will not be the same as at their nominal voltage because the power is less, so operating temperature is less, and temperature affects the resistance (by something like 7% between cold and nominal operating temperature).
Also we have to assume the loads are passive - nether the heater nor the bulb have any feedback to keep it emitting the same power if the supply changes. Non-dimmable LED bulbs will try to maintain the same current - and hence brightness - if the voltage changes. 71
Many heaters have a thermostat so average the same power output over several minutes if the voltage changes as they maintain the same temperature; if the question mentioned these cases then P would be the constant and you would be using P=IV all the way. A good answer should mention the assumptions being made.
You can either substitute I=V/R into P=VI, giving you your P=V²/R which gives you R=V²/P, nominal resistance of the loads from nominal power at specified voltage, or do it in two steps to get the current at specified voltage and nominal power then I=V/R for the nominal resistance.
You could also approximate and assume that the current through the loads is the same as when the loads are at nominal power and voltage, and I=P_nominal/V_nominal .
Using P=VI this way is usually good enough for the real world - a 60W load is 0.5A, a 240W load 2A, so you want a wire which can carry 2.5A without dropping too much voltage. 2.5A through 6R gives 15V. That's more of a drop than you would tolerate in reality, and the approximation is worse the more the drop is. But the approximation is good enough if you're adding up the load on an extension lead or a housing circuit, as you would size the wire for only a few percent drop.
For passive loads, this error would mean you overestimate the current so it is a safe approximation. For something like a switched mode supply for a servo motor, the motor has the same power requirement whatever the input to the supply, so the supply has feedback to draw more current given a lower input voltage, and it wouldn't be a safe approximation - you'd have to solve a quadratic if you have a fixed power load fed by a fixed wire resistance ( V_load * I = P_load; V_load + I × R_wire = V_supply => V_load + P_load/V_load × R_wire = V_supply ).
Anyway, you shouldn't connect heaters to lighting circuits, and the I you get from P_nominal = V_nominal × I_nominial isn't the same value that would be in the lead wires because of the voltage drop, so applying P=VI when none of the terms are fixed doesn't give you anything useful. 71