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Consider the following :
1) The LM339 has an open collector output, so needs a pull-up resistor.
2) FETs need to transition rapidly between the on and off states, otherwise they heat up.
3) The gate capacitance of the FET has to be charged and discharged, which takes time. The time should be minimised by using low-resistance driver circuits.
4) The gate turn-on threshold, Vto, for your chosen FET is 2.9V. At that voltage the FET will conduct only microamps. It needs a gate voltage significantly higher to conduct significant current. 5V may not be high enough.
5) R2 and R3 have impractically high values. You could reduce them by two or three orders of magnitude.
Your schematic is hard to read because of all the zeros. Instead of 0.000005, for example, you could write 5μ or 5u.