I have more than IZK current running through a Zener diode, but the voltage drop is below its Zener voltage, what does this mean?
Here is your effective circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
The voltage at the emitter of Q1 is going to be about 2.4V (the typical forward voltage of many red LEDs) and the voltage at the base of Q1 will be about 0.7V above that (so about 3.1V). Note that the 10K resistor is not doing anything meaningful.
So there would be about 3.1V across the zener, implying 3.9V across 200 ohms = 19.5mA, the majority of which is going into the base of Q1.
The zener voltage simply cannot exceed those limits provided something does not short out (which it will). In this circuit, because the current is not limited in any way, a large amount of current will flow. Eventually it will apply about 3.5V across the LED directly (assuming Q1 does not short out), which would be destructive.
If you wish to simply turn on your LED, there are simpler ways without a zener.