It's easier to understand with a drawing:
simulate this circuit – Schematic created using CircuitLab
The way you propose to use the NMOS as a high side switch is correct, you have to take the gate voltage above the supply voltage.
Let's use Vctrl = 10 V to open the NMOS and the supply is 50 V, that means there must be 60 V at the gate. Note that that is 60 V referenced to ground.
That -20 V / + 20 V limitation of the NMOS is for the NMOS itself, for its Vgs.
That Vgs is the voltage across the Gate and Source of the NMOS. Since the source of the NMOS is at 50 V when the NMOS is switched on, the Vgs will be 10 V, not 60 V. So this condition is allowed and OK.
But watch out! If you now want to switch the NMOS off you might have a problem.
Let's say you make Vctrl = 0 V then the gate will be at 50 V. That will not turn off the NMOS will as the voltage at the source will drop until it opens the so much that the NMOS will still conduct somewhat. It will work as a source follower. Avoid that ! What needs to be done is to make the Vgs of the NMOS itself equal to zero.
In order to do that you need to take the Gate all the way down to 0 Volt so ground. That means 0 Volt referenced to ground, just making Vgs = 0 is not enough as explained above. For this we would need to make Vctrl = -50 V !