Electronic – MOSFET Over Heating. So hot it’s melting the solder

Short answer: You're not giving it enough gate voltage for it to turn on completely. It needs 10V for the guaranteed Rds(on) of 40m\$\Omega\$.

The BUZ11 is a fine MOSFET, but it is not designed to operate from a 3.3V drive voltage. With 3.3V on the gate (relative to source) it will only turn on partially, and will begin to heat rapidly with a heavy load. The threshold voltage has a negative temperature coefficient, so the hotter it gets, the more it tends to turn on. Needless to say, this is not a good way to operate the MOSFET.

You could either get a MOSFET that has a guaranteed on-resistance with a 3.3V or lower drive voltage, or boost the gate drive voltage to allow you to use a part like the BUZ11.

There may not be any MOSFETs that are as good as the BUZ11 with 3.3V drive (especially the 50V voltage rating), at any price. If you can reduce the voltage rating to 20V, there may be a number of them available- but in TO-220, I see only the IRL3103PBF, which is probably okay but not guaranteed at 3.3V drive.

You could simply use a transistor to drive the MOSFET gate as so:

This would also tend to protect your Raspberry pi from some types of failure and bad connections.. a gate to drain short on a direct-driven MOSFET would otherwise release the smoke.

Note that "high" = "off" since the BJT acts as an inverter. If this is undesirable, you can add another BJT inverter.

A circuit diagram of your setup, particularly including your MOSFET driver would be really nice. However, I'll go out on a limb here and suggest that you need to look closely at your drive circuits. It is entirely possible that you are driving your MOSFET gates from 3.3 volt logic. This is perfectly doable - as long as you are using MOSFETs with logic-level gate thresholds. A lot of power MOSFETs need a minimum of 4 volts on the gate to ensure full turn-on. If you're only giving it 3.3 volts, it will only turn on partially, and will dissipate too much power. It's important that you realize that operating a transistor within both voltage and current limits can still kill it if you don't get rid of the heat dissipated, and at high currents it's easy to generate too much heat.

There is a quick check for this (if you want physical proof): Get ready to sacrifice one more MOSFET, but hey, who's counting, right? Drive a heating element full on. Quick like a bunny, measure the voltage across the nichrome and the voltage across the MOSFET. If your MOSFET voltage is not less than about 10% of the nichrome voltage, you're doing something wrong, and less is better in this case. You don't mention your drive voltage, but it has to be less than 25 volts. Let's say it's 20 volts, and let's say the current is 10 amps - this is just to illustrate, OK? Then total power dissipated is 200 watts, and the effective resistance of the total load is 2 ohms. If your MOSFET is fully on, I'd expect an Rds of .1 ohms or less (and this will give a MOSFET voltage about 5% of the nichrome voltage). This would provide a MOSFET dissipation of 10 watts. Without a heat sink, this would kill the MOSFET, so you need a heat sink in any case. And this better not be one of those little U-shaped jobbers, either. You need real heat sink, possibly with a fan. With more airflow you can use a smaller heat sink.

You need to consult the data sheet for your MOSFET to determine both Vgs(th),the threshold gate voltage, and Rds(on), the on-resistance when the gate is properly driven. Then you will need the specs on your heat sink, specifically the thermal resistance to ambient. You will also need to do a little research on how to specify a heat sink.

As for some of your other questions, consider the two voltages you measured. If they both add up to the nominal voltage of your drive converter, you are not drawing too much current, and you don't have a short. You are just fatally abusing your MOSFETs.