Opening Summary
The circuit is the following (you really should learn to redraw schematics for better reading):
simulate this circuit – Schematic created using CircuitLab
If you are willing to assume that the series current through \$R\$ and the zener diode is very large with respect to the base current required by \$Q_1\$, then it follows that the voltage at the base of \$Q_1\$ is fixed and determined by the zener diode. So we can ignore the details of the behavior of the left-hand side and just claim that \$Q_1\$'s base voltage is fixed for the purposes of analysis.
That leaves the right-hand side to worry over.
Solving for \$V_{_\text{OUT}}\$
For simplification purposes, we can assume that the emitter and collector currents are equal. (They aren't. But for reasonable values of \$\beta\$ they are close enough that we can consider them equal.) And we can ignore the -1 term in the Shockley diode model to get:
$$V_{_\text{OUT}}\approx R_{_\text{LOAD}}\,I_{_\text{C}}=R_{_\text{LOAD}}\,I_{_\text{SAT}}\,e^{^{\left[\frac{V_{_\text{Z}}-V_{_\text{OUT}}}{V_T}\right]}}$$
This solves out as:
$$V_{_\text{OUT}}\approx V_T\cdot \operatorname{LambertW}\left(\frac{I_{_\text{SAT}}\,R_{_\text{LOAD}}}{V_T}\cdot e^{^{\left[\frac{V_{_\text{Z}}}{V_T}\right]}}\right)$$
We know that \$V_T\approx \frac1{40}\$ so it follows that \$\operatorname{LambertW}\left(\frac{I_{_\text{SAT}}\,R_{_\text{LOAD}}}{V_T}\cdot e^{^{\left[\frac{V_{_\text{Z}}}{V_T}\right]}}\right)\approx 40\,V_{_\text{OUT}}\$.
Sensitivity
The concept of a sensitivity equation starts with recognizing that a %-change in something (using finite values) would be \$\frac{\Delta x}{x}\$. But using infinitesimal values this is still more precisely stated as \$\frac{\text{d}\, x}{x}\$. A sensitivity equation relates the %-change of one thing to the %-change of another thing, so the expression might look like \$\frac{\frac{\text{d}\, y}{y}}{\frac{\text{d}\, x}{x}}=\frac{\text{d}\, y}{\text{d}\, x}\cdot\frac{x}{y}\$, which gives you the %-change of y as compared to the %-change of x.
In this case, we may want to know what %-change in \$V_{_\text{OUT}}\$ results from some given %-change in \$R_{_\text{LOAD}}\$. Performing the required calculus, this sensitivity equation is then:
$$\frac{\% V_{_\text{OUT}}}{\% R_{_\text{LOAD}}}=\frac{V_T}{V_{_\text{OUT}}\left(1+\frac1{\operatorname{LambertW}\left(\frac{I_{_\text{SAT}}\,R_{_\text{LOAD}}}{V_T}\,\cdot\, e^{^{\left[\frac{V_{_\text{Z}}}{V_T}\right]}}\right)}\right)}$$
But we already determined that the LambertW function is going to be rather large (about \$40\cdot V_{_\text{OUT}}\$.) So the last term in the denominator therefore fades to zero.
The simplified result is then:
$$\frac{\% V_{_\text{OUT}}}{\% R_{_\text{LOAD}}}=\frac{V_T}{V_{_\text{OUT}}}$$
Or, in more useful terms:
$$\% V_{_\text{OUT}}=\left[\frac{V_T}{V_{_\text{OUT}}}\right]\,\% R_{_\text{LOAD}}$$
Final Summary
Roughly speaking this means, taking everything into account, we should find that the %-change in \$V_{_\text{OUT}}\$ will be about \$\frac1{40\,\cdot\, V_{_\text{OUT}}}\$ of the %-change in \$R_{_\text{LOAD}}\$.
This pretty much tells you that the %-change in \$V_{_\text{OUT}}\$ is very, very, very much less than the %-change in \$R_{_\text{LOAD}}\$ for all practical circumstances. (Only a few parts per thousand.)
Or, put again in other more qualitative terms, the output voltage won't budge much with respect to load changes.
And yes, stuff like this can be quantified (as shown above) and not just hand-waved-at. You want numbers? You got numbers. But when you get done with all that quantifying work, you'll once again find (in this case) that qualitative hand-waving is good enough for all intents and purposes.
Final Note: All of the above work assumes that \$V_{_\text{Z}}\$ is rock solid and doesn't itself move in response to load changes. The fact is, it does change because the base current is a function of the load current and a zener diode is specified for some operating current that is often quite limited. The above is good enough where the zener current is, as earlier stated, very much greater than the worst case expected base current. But it fails when that is no longer true and then a deeper analysis is required. Just keep it in mind.
