The partition function $\mathcal Z$ is the sum over all those these exponential indexed by the state index. In German it's called "$\mathcal Z$ustandssumme", literally "state sum".
You could device a scheme of indexing the possible overall energies so that you could write it as $\sum_n$, i.e. finding an injection ${\mathbb N}\to{\mathbb N}^2$, but if the thing converges nicely, the result is the same as if you write $\sum_{n_1}\sum_{n_2}$.
I'm not sure I will be able to clarify all your doubts, but this is the right approach to tackle this problem.
We consider $N_s$ available adsorption sites, an energy $\epsilon$ for each bound state, chemical potential $\mu$ and temperature $T$.
The grandpartition function $\mathcal Q$ is always expressed as
$$
\mathcal Q= \sum_{N=0}^\infty e^{\beta \mu N} Z_N
$$
in terms of the $N$-particle partition function. The latter is defined by
$$
Z_N=\sum_{\substack{N-\text{particle}\\ \text{states}}} e^{-\beta E(\text{state})}\,.
$$
In our case, the energy of a given $N$-particle state, of the ensemble of bound states, is $N\epsilon$ and there are $\binom{N_s}{N}$ such $N$-particle states, since each of the $N$ bound states can be placed by choosing one site among the $N_s$ sites, without repetition:
$$
Z_N=\binom{N_s}{N}e^{-\beta \epsilon N}\,.
$$
Note that this partition function does not bear a factorized form.
Finally,
$$
\mathcal Q= \sum_{N=0}^\infty \binom{N_s}{N}e^{\beta (\mu-\epsilon) N}
= \sum_{N=0}^{N_s} \binom{N_s}{N}e^{\beta (\mu-\epsilon) N}
=(1+e^{\beta(\mu-\epsilon)})^{N_s}\,,
$$
where in the last step we used the binomial formula
$$
(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k.
$$
EDIT: One can also reason directly using the grandpartition function as follows. Using the occupation number representation $\{n_{\alpha,k}\}$ for noninteracting particles, with $|\alpha, k\rangle$ labelling single-particle states with energy $\epsilon_\alpha$ and $g_\alpha$-fold degeneracy $k=1,2,\ldots,g_{\alpha}$,
$$
\mathcal Q= \sum_{N=0}^\infty e^{\beta\mu N}
\sum_{\substack{
\{n_{\alpha, k}\}:\\
\sum_{\alpha, k}n_{\alpha,k}=N}}
e^{-\beta \sum_{\alpha, k}n_{\alpha,k} \epsilon_\alpha }
=
\sum_{\{n_{\alpha,k}\}} e^{\beta \sum_{\alpha, k}n_{\alpha,k} (\mu-\epsilon_\alpha) } = \prod_{\alpha,k} \sum_{n_{\alpha, k}}e^{\beta(\mu-\epsilon_\alpha)n_{\alpha,k}}\,.
$$
This is a proof that, for noninteracting systems, the grandpartition function always takes the factorized form
$$
\mathcal Q = \prod_{\alpha, k} \left( \sum_{n_{\alpha,k}} e^{\beta(\mu-\epsilon_\alpha)n_{\alpha,k}}\right)\,.
$$
In the case at hand, the single-particle states all have energy $\epsilon$ and have multiplicity $N_s$; in the above notation $\alpha=1$ and $k=1,2,\ldots,N_s$ so
$$
\mathcal Q = \prod_{k=1}^{N_s}(1+e^{\beta(\mu-\epsilon)})=(1+e^{\beta(\mu-\epsilon)})^{N_s}\,.
$$
Best Answer
Here's my shot at it and my whole thought process so we are checking each other.
If we put in the Hamiltonian into the exponent and talk it through, we get the following: $$ \int e^{-\beta( \sum_i p^{2}_i/2m_1 +\sum_j p^{2}_j/2m_2)} $$ but what is the measure of integration? well they are non interacting, so we have to look at each individual particle, of each type, and then sum over all the possible momenta they could have $$ \int e^{-\beta( \sum_i p^{2}_i/2m_1 +\sum_j p^{2}_j/2m_2)}\Pi_i d^3p_i \Pi_j d^3p_j $$ where I have ignored anything to do with space recklessly. now my guess for the density of states part is $$ \frac{d^3p d^3 x}{\hbar^3}=\frac{4\pi V p^2 dp}{\hbar^3} $$ (there is no spatial dependence, so I just integrated over the whole volume) so really I had $$ \int e^{-\beta H}\frac{d^{3n}p d^{3n} x}{\hbar^{3n}} $$ and I need to put that in $$ \frac{(4\pi)^{N_1 +N_2}V^{N_1+N_2}}{\hbar^{3(N_1+N_2)}} \int_{0}^{\infty} e^{-\beta( \sum_i p^{2}_i/2m_1 +\sum_j p^{2}_j/2m_2)}(\Pi_i p^{2}_i dp_i)( \Pi_j p^{2}_j dp_j) $$ This is a ton of do-able individual integrals, all multiplied, because the sum in the exponents are multiplied down below. I have been really sloppy about numerical factors (like $e^{-\beta N_2 \Delta}$) and constants, sorry. I hope this helps/is correct.