I understand why some transistors have a built-in (normal) diode from drain to source, so that when the transistor turns off, back-emf can flow back up, specially important in inductive charges, but why do some MOSFET like IRF540 have a zener? From what I understand, a zener diode conducts backwards with a specified drop voltage. I can only suspect that this zener has something to do with the absolute maximum voltage sustained between source and drain, specified in the datasheet, is that correct?
Also, I am trying to build this circuit: 
But the transistor I am using as Q1 is currently IRF540 that has a zener diode instead of a normal one. As expected, it doesn't work very well. As proof of concept, with a normal diode (no Q1), Q2 I can successfully charge the gate and it lets current pass through.
I tried to put a normal diode in parallel with the intrinsic zener and it still doesn't work...
Answer
I think you've got it mixed up. The body diode is a part of the structure of the MOSFET and does not help when switching an inductive load with a single MOSFET.
The zener in the symbol represents the avalanche rating of the MOSFET and is of use when switching an unclamped inductive load. You won't see it conduct unless you exceed the voltage rating \$V_{DS}\$ and get to \$V_{(BR)DSS}\$.
If you are actually getting +/-12V out of the pulse transformer, I don't see any obvious reason why that part would not work. Your negative swing out of the pulse transformer does have to be sufficiently large to turn the control MOSFET on, when loaded with both gate charges (-10V would be good).
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