I don't really see the answer in the other answer so let me do the calculation here. Your general Boltzmann Ansatz says that the probability of a state $n$ depends on its energy as
$$ p_n = C \exp(-\beta E_n) $$
where $\beta = 1/kT$. Fermions are identical particles that, for each "box" or one-particle state they can occupy (given e.g. by $nlms$ in the case of the Hydrogen atom-like states), admit either $N=0$ or $N=1$ particles in it. Higher numbers are forbidden by the Pauli exclusion principle. The energies of the multi-particle state with $N=1$ and $N=0$ in a particular one-particle state $nlms$ differ by $\epsilon$. Consequently,
$$ \frac{p_1}{p_0} = \frac{C\exp(-\beta (E+\epsilon))}{\exp(-\beta E)} = \exp(-\beta \epsilon) $$
where I used the Boltzmann distribution. However, the probabilities that the number of particles in the given one-particle state is equal to $N=0$ or $N=1$ must add to one,
$$ p_0 + p_1 = 1.$$
These conditions are obviously solved by
$$ p_0 = \frac{1}{1+\exp(-\beta\epsilon)}, \qquad p_1 = \frac{\exp(-\beta\epsilon)}{1+\exp(-\beta\epsilon)}, $$
which implies that the expectation value of $n$ is equal to the right formula for the Fermi-Dirac distribution:
$$\langle N \rangle = p_0\times 0 + p_1 \times 1 = p_1= \frac{1}{\exp(\beta\epsilon)+1} $$
The calculation for bosons is analogous except that the Pauli exclusion principle doesn't restrict $N$. So the number of particles (indistinguishable bosons) in the given one-particle state may be $N=0,1,2,\dots $. For each such number $N$, we have exactly one distinct state (because we can't distinguish the particles). The probability of each such state is called $p_n$ where $n=0,1,2,\dots$.
We still have
$$\frac{p_{n+1}}{p_n} = \exp(-\beta\epsilon) $$
and
$$ p_0 + p_1 + p_2 + \dots = 1 $$
These conditions are solved by
$$ p_n = \frac{\exp(-n\beta\epsilon)}{1+\exp(-\beta\epsilon)+\exp(-2\beta\epsilon)+\dots } $$
Note that the ratio of the adjacent $p_n$ is what it should be and the denominator was chosen so that all the $p_n$ from $n=0,1,2\dots$ sum up to one.
The expectation value of the number of particles is
$$ \langle N \rangle = p_0 \times 0 + p_1 \times 1 + p_2\times 2 + \dots $$
because the number of particles, an integer, must be weighted by the probability of each such possibility. The denominator is still inherited from the denominator of $p_n$ above; it is equal to a geometric series that sums up to
$$ \frac{1}{1-q} = \frac{1}{1-\exp(-\beta\epsilon)} $$
Don't forget that $1-\exp(-\beta\epsilon)$ is in the denominator of the denominator, so it is effectively in the numerator.
However, the numerator of $\langle N \rangle$ is tougher and contains the extra factor of $n$ in each term. Nevertheless, the sum is analytically calculable:
$$ \sum_{n=0}^\infty n \exp(-n \beta\epsilon) = - \frac{\partial}{\partial (\beta\epsilon)} \sum_{n=0}^\infty \exp(-n \beta\epsilon) =\dots$$
$$\dots = - \frac{\partial}{\partial (\beta\epsilon)} \frac{1}{1-\exp(-\beta\epsilon)} = \frac{\exp(-\beta\epsilon)}{(1-\exp(-\beta\epsilon))^2} $$
This result's denominator has a second power. One of the copies gets cancelled with the denominator before and the result is therefore
$$ \langle N \rangle = \frac{\exp(-\beta\epsilon)}{1-\exp(-\beta\epsilon)} = \frac{1}{\exp(\beta\epsilon)-1} $$
which is the Bose-Einstein distribution.
You could also obtain another version of the Boltzmann distribution for distinguishable particles by a similar calculation. For such particles, $N$ could take the same values as it did for bosons. However, the multiparticle state with $N$ particles in the one-particle state would be degenerate because the particles are distinguishable. There would actually be $N!$ multiparticle states with $N$ particles in them. The sum would yield a Taylor expansion for the same exponential.
Note added later: the derivation above was for $\mu=0$. When the chemical potential is nonzero, all appearances of $\epsilon$ have to be replaced by $(\epsilon-\mu)$. Of course that one may only talk about a well-defined value of $\mu$ when we deal with a grand canonical potential; it is impossible to derive a formula depending on $\mu$ from one that contains no $\mu$ and assumes it's ill-defined. The derivation above was meant to show that the difficult $1/(\exp\pm 1)$ structures do appear from a simpler $\exp(-\beta E)$ Ansatz because I think it's the only nontrivial thing to be shown while discussing the relations between the Boltzmann and BE/FD distributions. If that derivation proves the same link as the textbook does, then I apologize but I think there is "nothing else" of a similar kind to be proven.
The difference lies in the way we count the number of states of the system in quantum and classical cases.
The formulas you wrote are actually for the grand canonical partition functions for a single energy state, not for the whole system including all the energy states. The total grand canonical partition function is $$\mathcal{Z} = \sum_{all\ states}{e^{-\beta(E-N\mu)}} = \sum_{N=0}^\infty\sum_{\{E\}}{e^{-\beta(E-N\mu)}}$$
Now, if the particles are bosons, then the energy eigenstates are countable as $\{\epsilon_i\}$ and $\mathcal{Z}$ would be $$\mathcal{Z} = \sum_{\{n_i\}}e^{-\beta\sum_{i}{n_i(\epsilon_i-\mu)}} = \sum_{\{n_i\}}\prod_ie^{-\beta n_i(\epsilon_i-\mu)}=\prod_i \mathcal{Z_i}^{B-E}$$
where $n_i$ is the number of particles in $i$-th energy state, thus $\sum_i{n_i}=N$, and
$$\mathcal{Z_i}^{B-E} = \sum_{n=0}^\infty e^{-n\beta(\epsilon_i-\mu)}$$
Here, $\mathcal{Z_i}^{B-E}$ is the grand canonical partition function for one energy eigenstate with energy $\epsilon_i$ in Bose-Einstein statistics.
On the other hand, in classical regime the energy of a particle can take any energy. In this case, one point in the $6N$-dimensional phase space denotes one state of system. Therefore, the energy states are not countable as there is an infinite number of points in the phase space within any phase space volume. To count the states we take the "semi-classical" approach by taking the phase space volume of one state of the system to be $(2\pi\hbar)^{3N}$. We can then integrate over the whole phase space and divide the integral by this unit volume to get the number of states. However, as the particles are assumed to be indistinguishable, any permutation of the system configuration (the set of $\{\vec{x}_n,\ \vec{p}_n\}$) would actually be the same state of the system. Therefore, when we integrate over the whole phase space volume we overcount the total number of states by $N!$. That's why we need to divide the integral by Gibbs factor $N!$. For a system of non-interacting particles, the N-particle canonical partition function then can be written as $Z_N = \frac{Z_1^N}{N!}$ where $Z_1$ is the canonical partition function for one particle.
Now, the grand canonical partition function for a classical system would be
$$\begin{align}
\mathcal{Z}&=\sum_{N=0}^\infty{\int_0^\infty dE\ \Omega(E,N)e^{-\beta(E-\mu N)}}
=\sum_{N=0}^\infty e^{\beta\mu N} Z_N = \sum_{N=0}^\infty e^{\beta\mu N} \frac{Z_1^N}{N!}\\
\end{align}$$
If we want to derive the partition function $\mathcal{Z_i}$ for a single energy state similar to the Bose-Einstein statistics, we can assume the energy of a single particle to be discrete and countable as $\{\epsilon_i\}$. This can be achieved by dividing the one particle phase space into s of unit volume and assigning one representative energy to every unit volume section. Then, the grand canonical partition function is,
$$ \begin{align}
\mathcal{Z} &= \sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} (\sum_i e^{-\beta\epsilon_i})^N\\
&=\sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} \sum_{\{n_i\},\sum n_i = N} \frac{N!}{\prod_i n_i!} e^{-\beta\sum_i n_i\epsilon_i}\\
&=\sum_{\{n_i\}}\prod_i \frac{1}{n_i!} e^{-\beta n_i(\epsilon_i-\mu)}\\
&=\prod_i \sum_{n=0}^\infty \frac{1}{n!} e^{-\beta n(\epsilon_i-\mu)} = \prod_i \mathcal{Z_i}^{M-B}
\end{align}$$
This $\mathcal{Z_i}^{M-B}$ is the single energy state grand canonical partition function in Maxwell-Boltzmann statistics.
Maxwell-Boltzmann statistics is the classical limit for Bose-Einstein statistics. The condition for the system to be classical is the single state occupation number $\bar{n}$ to satisfy $\bar{n} \ll 1$, in other words, the total number of single particle states $M$ should satisfy $N \ll M$. As, the single particle partition function $Z_1$ is actually a weighted sum over all the states, $N\ll Z_1$ will satisfy $N \ll M$ for the system to be classical.
Best Answer
I have just found a way to prove that $\ln Z = \frac{PV}{kT}$ as follows: the grand canonical potential is defined, in thermodynamics, to be $ \Xi = U - TS - \mu N $. Using Euler relation (Callen, eq. (3.6), also known as Euler integrals here in Wikipedia) $ U = TS - PV + \mu N $ then $$ \Xi = -PV. $$
On the other hand the grand canonical potential can be obtained from the grand canonical partition function as $$ \Xi = -kT \ln Z .$$
Now it is trivial that $$ \ln Z = \frac{PV}{kT} $$