I understand what you suggest, but honestly I don't yet understand how it works.
M1 forms a cascode with the internal switching transistor of U1. When U1's output transistor is off, the voltage on M1's source is just below the gate voltage and only a tiny leakage current flows. When U1's output transistor turns on, M1 also turns hard on, as its source voltage is now well below, the gate voltage. Because M1 forms a source follower, when U1's output transistor is off, the voltage across the output transistor always stays below 12V.
Let's assume we use 12V as input voltage. What would be the maximum output voltage (if the diode is exchanged)?
The maximum ratings of M1 + 12V minus the gate threshold voltage, at whatever the tiny leakage current though U1's output transistor is, minus the diode's forward voltage drop, or the diode's maximum reverse voltage rating, whichever is lower.
The RVQ040N05 has a maximum drain source voltage rating of 45V. The gate threshold is specified to be between 1V and 2.5V, with a drain current of 1mA and at the tiny leakage current through U1's output transistor, it'll be even lower. It's probably safe to say the maximum output voltage will be around 54V, with this arrangement.
EDIT:
I forgot that M1 will also leak a bit of current, so its source voltage will be a bit higher than I initially thought. I doubt it's an issue, even if it does exceed the maximum voltage rating of U1's output transistor, because the current will be limited to such a low level. A high value gate-source resistor, say 10M would keep the source at 12V, if you're paranoid.