quantum mechanics - Where is quantization used in deriving Planck's law?

There are several explanations for how Planck used quantization to explain blackbody radiation correctly without the ultraviolet catastrophe. I will follow this explanation.

For a cavity, the mode density per volume is $8\pi\nu^2/c^3$. Classically, the probability of occupation is equal for all modes and each mode has an energy $kT$. Thus, one obtains the Rayleigh-Jeans formula for blackbody radiation

$$I = \frac{8\pi\nu^2}{c^3}kT \, .$$

This blows up for large $\nu$. Planck's law instead says that the probability of occupation of a mode is $$\frac{1}{e^{h\nu/kT} - 1}$$ and the energy associated with that mode is $h\nu$. Therefore, we have $$I = \frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT} - 1} \, .$$ This equation behaves correctly for large $\nu$. Where exactly did one need quantization?

For instance, I understand that the Boltzmann distribution (which works for continuous distributions) has probability of occupation that goes as $p(E) = e^{-E/kT}$. Let energy be proportional to frequency (but not quantized) and this also averts the ultraviolet catastrophe and qualitatively produces the same shape as Planck's law.

EDIT

Nope, the above statement is wrong. I realized that if you actually do the integration correctly, you get the law below which still blows up for large $\mu$ $$I = \frac{8\pi\nu^2}{c^3}kT \, .$$ END OF EDIT

So where does the requirement of quantization exactly come in?

This question is closely related to this one and this one but is not a duplicate since those questions were asked at a more basic level.

Answer

From the classical Boltzmann theory you have the probability that the mode has energy $E$ $$ p(E) = A e^{-E/kT} \tag{1} $$ (with an $A$ so that the total probability is $\int_0^\infty p(E)\mathrm{d}E = 1$).
From this distribution you get the average energy $$ \bar{E} = \int_0^\infty p(E) E \mathrm{d}E = kT \tag{2} $$ This leads to the Raleigh-Jeans spectrum, but unfortunately it does not reproduce the experimentally measured black-body radiation.

Planck found that instead of equation (2) the average energy of a mode needs to be $$ \bar{E} = \frac{h\nu}{e^{h\nu/kT}-1} \tag{3}$$ in order to correctly reproduce the experimentally measured black-body radiation. By curve-fitting he also found the numerical value of $h = 6.6 \cdot 10^{-34} \text{Js}$.

There was no physical explanation for equation (3) available until this time. But Planck saw (you may call it mathematical intuition) that $$ p_n = A e^{-nh\nu/kT}, \text{with } n = 0, 1, 2, \dots \infty \tag{4} $$ (with an $A$ so that the total probability is $\sum_{n=0}^\infty p_n = 1$) would give the correct average energy (3). $$ \bar{E} = \sum_{n=0}^\infty p_n nh\nu = \frac{h\nu}{e^{h\nu/kT}-1} \tag{5}$$ You can find the details of proving equation (5) from equation (4) for example in The Derivation of the Planck formula (pages 9-10).

Now, in equations (4) and (5) $n$ and $E_n = nh\nu$ obviously have discrete values, in contrast to $E$ in equations (1) and (2) having continuous values. The physical interpretation of this is that the mode doesn't have arbitrary energies $E$, but only discrete energies $E_n = nh\nu$. Or saying it more vividly: Light of frequency $\nu$ consists of particles, each having energy $h\nu$.