There is no capacitance across the input supply to the board. There are in other places but not directly at the inputs of the power leads and they are actually something like 1.5m each in length.
Then this cannot work at all, except with pure luck. Input power bypassing with caps is an absolute must in any electronic product using semiconductors, no matter how simple case of switching it appears.
Both sides are as important; the inductance caused by 1.5m leads on the input side is as problematic as it would be on the output side. In the output side, you deal with it using the freewheeling diode; in the input side, you need bypass caps to make the loop smaller, to decrease the inductance.
If we assume 2uH of inductance on the input leads (approximated using
https://www.eeweb.com/toolbox/wire-inductance/ , pair of lines, 1mm dia, 10mm spacing), resistance-limited current of 33V/6ohm = 5.5A, then we are storing quite a lot of energy in that inductance. The voltage WILL overshoot quite a bit. I can't come up with a simple way of saying how much exactly, but you could easily simulate it in LTspice.
Assuming that the culprit is the current, you measured diode current, but you didn't measure the DC link voltage, did you?
I'm not saying this is the definite answer, but this is the most prominent culprit to me. Supply bypassing is an absolute must when switching anything with MOSFETs, unless you really slow down the switching.