Here's what happens when you turn off the switch signal: -

Initially, after switching the control signal off, stored energy in the inductor is gradually released into the zener diode via D2. This causes a voltage peak above the power rail. Theoretically, with a 10 volt zener and 0.7 volts from D2, that voltage peak should be about 21 volts i.e. 10.7 volts plus the rail voltage of 10 volts.
However, the inductor current will be about 10 amps (due to \$V = L\frac{di}{dt}\$) and your dt value of 1 milli second mentioned in your spice circuit. That initial current into the BZX84C10 would almost certainly destroy it but, the simulation does what it can and allows the initial peak of voltage to be much higher than that implied by just the 10 volt rating of the zener.
Once the stored energy has depleted somewhat, the zener voltage cannot be sustained any more and that is when the zener no longer operates as a clamp but returns to being just parasitic capacitance (a few pF to a few tens of pF).
And now, this is the final burn-off of remaining energy stored in the inductor - all the circuit allows it to do is ring until decayed to zero.