dueling pullups/pulldowns
A dueling pull-up and pull-down just give you a voltage divider. In this case, with 1.1k on top and 10k on bottom, you get 90% of your source voltage or 4.5V.
Looking at this MOSFET's datasheet, it's fully on at 4.5V and they even specify the on-resistance of 3.0mOhm at 4.5V, so that's actually plenty. If you wanted to get closer to your full 5V, you could change R31 to 100k for ~4.95V. All R31 does is make sure the gate capacitance of the MOSFET discharges when you aren't driving it, so that it turns off when there's no control signal to it. FETs don't turn off until you discharge their gate.
That said, I don't think this is the circuit you want. With an open drain, you not only have very little drive current to turn the FET on, but also with no input to the buffer, the FET will be on(as soon as you power your circuit and before the microcontroller takes control).
Ideally you want an output stage with a pair of transistors that can push/pull the gate. This FET doesn't have a huge gate and I imagine you aren't switching it very often, so that can be almost anything. I attached a simple drive circuit with typical resistor values. I don't guarantee they're ideal values but they're safe values for the microcontroller and drive transistors, so the MCU can't push more than 3mA and the transistors can't push more than 500mA.
If you're trying to drive a lot more than the 10A I've shown through the heater, like the 100A that FET is rated for, you might want to just use a dedicated gate drive IC, or at least use FETs as your drive transistors, and reduce the gate resistance to 1-2ohms, so that your switching times are as short as possible to limit switching loss.