So if you are just turning an LED on and off with 20mA, I know I would use an NPN transistor (2N3904 or 2N2222 type). But a MOSFET will certainly do the job just as well. The BSS138 as a low side switch will work fine for an LED. I do it all the time as do many other designers.
However, if you are enabling power to other circuitry, low side switching gets very tricky, because once you do low side switching you no longer have a single ground plane. I strongly recommend you avoid low side switching if you are trying to control power to other IC's that you are communicating with.
In your case, since you have a 5V VCC available to switch a 3.3V rail, you actually can use a high-side NMOS without any special additional circuitry. But you need to find one that will fully turn on with Vgs of 1.7V. And remember, 1.7V is the nominal voltage. If your 3.3V rail is running a little high and your 5V rail is running a little low, you can find yourself with a lot less than 1.7V left to turn on the FET.
The Vgs(th) rating of the FET is not what you should be looking at when you are using the FET as a switch. You should note what is the lowest voltage at which Rds(on) is specified. In your case, you want a low Rds(on) to be listed in the datasheet when Vgs is 1.5V or even less if possible.
Purely as an example (shopping questions are off-topic here) I did find one such FET that might work for you. It is the Diodes Inc part number DMN1019USN-13. Rds(on) is specified as 41mOhms max at Vgs=1.2V. So that should work for you.
A word of caution: for MOSFET's with such low switching thresholds, you need to make sure that you have it turned off when you want it off. Any little voltage may actually cause the channel to start conducting a bit. In your case that should be no problem because when you drive the gate to ground, Vgs will be negative. But others who may be reading this may be in a slightly different situation.