Since you're just switching a magnet, you don't need high speed, or high gate current. However, since your MCU operates on 3.3 volts, you do need some sort of driver to guarantee that your MOSFET gets turned on strongly, and 3.3 volts on an output just won't do that.
For a simple case like this, all you need is a transistor and a few resistors, and
simulate this circuit – Schematic created using CircuitLab
this will do, as long as you don't mind the fact that the signal is inverted. That is, a high output at the MCU will turn the relay off, not on. Note a few things. R2 and R3, when the transistor is off, set the gate drive at 12 volts. Without the combination, using only R2 would apply a maximum of 24 volts to the gate, and this exceeds the maximum specification. Also, D1 is called a flyback diode, and you should always use one when switching anything with a coil. If you don't, when you turn off the relay you'll get a big voltage spike across it, and you may well kill your transistor. Worse, sometimes it will take multiple operations to kill the transistor, so you think you've got a working circuit, but you can't understand why it's unreliable. The diode should be rated for whatever current the relay coil draws, and have a higher voltage rating than the power supply.