The thing is that Maxwell-Boltzmann does hold for any system. Quantum particles, however, are not only identical, but completely indistinguishable. For such particles, the distribution is characterized solely by the occupation number.
Consider a system of noninteracting identical particles. Suppose the single particles have a known Hamiltonian $H$ with eigenstates $|k\rangle$, such that
$$H|k\rangle=E_k|k\rangle.$$
If we have a bunch of say, $N\gg1$ of these particles, we write the quantum state of the whole system in a different basis (the Fock basis), given by the number of particles in state $k, n_k$:
$$|n_1,n_2,...\rangle,$$
And this basis completely characterizes our quantum state. These are eigenstates of the many body Hamiltonian, with
$$H|n_1,n_2,...\rangle=\left(\sum\limits_kn_kE_k\right)|n_1,n_2,...\rangle.$$
Suppose then that our ensemble of particles has a state probability
$$p(n_1,n_2,...).$$
Maxwell-Boltzmann statistics is generated by maximizing the Gibbs Entropy:
$$
S=-\sum\limits_{n_1,n_2,...}p(n_1,n_2,...)\log p(n_1,n_2,...),$$
Where $\log$ is the natural logarithm. We can then do it subject to the constraints of normalization, total energy and total number of particles:
$$\sum\limits_{n_1,n_2,...}p(n_1,n_2,...)=1,\quad\sum\limits_{n_1,n_2,...}p(n_1,n_2,...)\sum\limits_kn_kE_k=U,\quad\sum\limits_{n_1,n_2,..}p(n_1,n_2,...)\sum\limits_kn_k=N.$$
The maximization condition then corresponds to (using Lagrange multipliers, the calculation is a little involved)
$$p(n_1,n_2,...)=\prod\limits_k\frac{e^{-\beta n_k(E_k-\mu)}}{\sum\limits_{n_k'}e^{-\beta n_k'(E_k-\mu)}},$$
where $\beta$ and $\mu$ are parameters corresponding to the inverse temperature, $(k_BT)^{-1}$ and chemical potential. If we take the marginal distribution for a single occupation number, say, $n_j$, by summing over all other possible values for all the other occupation numbers, we get the probability distribution for the occupation number in question:
$$p_j(n)=\frac{e^{-\beta n(E_j-\mu)}}{\sum\limits_{n'}e^{-\beta n'(E_j-\mu)}}.$$
The step further taken by Bose-Einstein and Fermi-Dirac statistics is now computing the average, or expected occupation number of the quantum state $j$:
$$f_j=\langle n_j\rangle=\sum\limits_nnp_j(n).$$
A nice computational trick is then writing $f_j$ as follows:
$$f_j=\frac{\sum\limits_nne^{-\beta n(E_j-\mu)}}{\sum\limits_{n'}e^{-\beta n'(E_j-\mu)}}$$
$$f_j=\frac{1}{E_j-\mu}\frac{\sum\limits_nn(E_j-\mu)e^{-\beta n(E_j-\mu)}}{\sum\limits_{n'}e^{-\beta n'(E_j-\mu)}}$$
$$f_j=-\frac{1}{E_j-\mu}\frac{\partial}{\partial\beta}\log\left(\sum\limits_ne^{-\beta n(E_j-\mu)}\right).$$
As such, if we can compute that logarithm on the right, we can find $f_j$. Here, we need to make an assumption.
Bosons:
For bosons, $n$ can be any integer from $0$ to $\infty$. As such, we write
$$\sum\limits_{n=0}^\infty e^{-\beta n(E_j-\mu)}=\frac{1}{1-e^{-\beta(E_j-\mu)}},$$
a convergent geometric series. We now plug this into our expression for $f_j$ and get
$$f_j=\frac{1}{e^{\beta(E_j-\mu)}-1},$$
which is Bose-Einstein statistics.
Fermions
For fermions, $n$ can be either 0 or 1. We then write the sum over $n$ as
$$1+e^{-\beta(E_j-\mu)},$$
Which then gives us
$$f_j=\frac{1}{e^{\beta(E_j-\mu)}+1},$$
The occupation number of Fermi-Dirac statistics.
EDIT
I just realized that I got ridiculously deep into the mathematics and forgot to anwer the question. The intuition behind all these calculations.
Thing is, in its essence, Maxwell-Boltzmann statistics is concerned with system states, and Bose-Einstein/Fermi-Dirac are concerned with particle states. It's easy to get both these concepts mixed up, but they are fundamentally different. While M-B gives us the probability for our system (all of our $N$ particles) to be in a specific configuration, B-E/F-D gives us the average number of particles that are in a specific state.
Since, by definition, these are averages, we have to average over all configurations of the system. And that is exactly what we did here.
It's also nice to mention that we can compute $\mu$ by solving the equation
$$\sum\limits_jf_j=N.$$
