First a zener diode is nothing but a diode, just a heavily doped one at that (Let's get to that later).
When you apply a voltage across a diode, you get an I-V characteristic like this
(Source:Google, Electronics notes)
What we need to focus on is the reverse bias part.
Suppose you apply a reverse biasing voltage. An electron crossing the depletion region (either in diffusion current or drift current) experiences a force towards the +ve side and a hole (assuming as an independent particle itself) towards the -ve side. In net effect this widens the depletion region, increasing the barrier voltage. This reduces the diffusion current( holes moving towards electrons and vice versa) to almost null but slightly increases the drift current (the hole-electron pair generated naturally by the semiconductor in the depletion region) which is still very low because of the small rate of generation of the hole electron pair.
(Source:Google, Khan Academy)
But if you keep increasing the voltage, then the electrons due to high speed collide with the electrons in the covalent bonds, breaking them, hence releasing a lot of electron-hole pairs suddenly.This is termed as avalanche- the sharp spike you see to the left.
Is this useful? On a first glimpse no, as the extremely high energy of electrons results in the diode melting.
However then comes zener diode. It is heavily doped, which causes the depletion region to be really small initially. As above , when you apply a reverse biasing voltage, it breaks down, but at a much lower voltage. So the electrons are not extremely energetic preventing damage to the diode.
The voltage supplied is used up as energy to break the covalent bonds. Also ideally it has no resistance. So it cannot consume more energy than this and hence you cannot increase the voltage across this beyond the breakdown voltage. I am refering to voltage across the zener diode, not across the source.
If the voltage across the source is low, such that you cannot have enough voltage across the zener diode to cause breakdown, then it has negligible current through it, which in practice is just like an open circuit.
However when sufficient voltage is given it has a constant voltage across it, and the rest of the voltage supplied by the source is dropped in the other resistors etc.
Now if we connect a load resistor across the zener diode in reverse breakdown, by kirchoff's voltage law it MUST have the same voltage dropped across it as the diode, which is a constant because as discussed above, for ideal zener diodes you just can't increase the voltage beyond that.
A major misconception in your question is that covalent bonds break when "Voltage across it is higher than the zener breakdown voltage" - It happens at the zener breakdown voltage