Showing that disjoint intervals implies independent number of arrivals for random point set If we have a number of i.i.d. random variables, $X_1, X_2, ..., X_z$ (where $z$ is the realisation of a random variable $Z\sim Po(\lambda)$ independent of each $X_i$) with pdf $f$ that form a random point set, then I want to show that for two disjoint sets (in this case intervals), we have that $N(A_1)$ and $N(A_2)$ are independent. These are the number of "arrivals" (as in a Poisson Process - but I haven't shown it to be one yet) in said intervals. I will denote $p_i:=\int_{A_i} f\,dx$ to be the probability of being in the interval $A_i$.
I want to show $P(N(A_1)=n\space\cap\space N(A_2)=m)=P(N(A_1)=n)P(N(A_2)=m)$. If we only consider the cases where $n+m\leq z$, then 
$$P(N(A_1)=n\space\cap\space N(A_2)=m)=P(N(A_1)=n)P(N(A_2)=m|N(A_1)=n)$$
So I want that $P(N(A_2)=m|N(A_1)=n)=P(N(A_2)=m)$. However, 
$$P(N(A_2)=m|N(A_1)=n)=\binom{z-n}{m}p_2^m(1-p_2)^{z-n-m}$$
is definitely not the same expression as the desired one, i.e. $\binom{z}{m}p_2^m(1-p_2)^{z-m}$. Where did I go wrong? I know for a fact that the set must describe a Poisson Process, so I must have made an error somewhere.

The point process you are considering is not a Poisson point process (PPP) but rather a binomial point process (BPP). Indeed, for a PPP, $N(A)$ is Poisson distributed while in your case $N(A)$ is a binomial random variable. In particular, even if $A_1$ and $A_2$ are disjoint, $N(A_1)$ and $N(A_2)$ are dependent since
$$N(A_1) + N(A_2) = N(A_1 \cup A_2) \leq z.$$
I should note however that there is an error in your calculation. We have
$$P(N(A_1) = n, \, N(A_2) = m) = \binom{z}{n} \binom{z-n}{m}p_1^np_2^m (1-p_1-p_2)^{z-n-m}\tag{1},$$
which is saying that we want $n$ points in $A_1$, $m$ points in $A_2$ AND $z-n-m$ points in $\mathbb{R}\setminus (A_1\cup A_2)$.
Final note: if you want a PPP then you should take $z$ Poisson distributed (instead of deterministic) and independent of the $X_i$.

EDIT: 
Let $Z$ be Poisson distributed with mean $\lambda$ and set $N = \sum_{i=1}^Z\delta_{X_i}$. Then
$$\begin{align}
P(N(A_1) = n, \, N(A_2) = m) &= \sum_{z=0}^\infty P(N(A_1) = n,\, N(A_2) = m, \, Z = z) \\
&= \sum_{z=n+m}^\infty P(N(A_1) = n,\, N(A_2) = m, \, Z = z)\\
&=\sum_{z=n+m}^\infty P(N(A_1) = n,\, N(A_2) = m| Z = z)e^{-\lambda}\frac{\lambda^z}{z!},
\end{align}$$
where  I used that the summand is zero if $z < n+m$ for the second equality and that $Z \sim Po(\lambda)$ for the last. Now, conditionally on $Z=z$, the computation above is valid and gives that
$$\begin{align}P(N(A_1) = n, \, N(A_2) = m|Z=z) &= \binom{z}{n} \binom{z-n}{m}p_1^np_2^m (1-p_1-p_2)^{z-n-m} \\
&= \frac{z!}{n! m!(z-n-m)!}p_1^n p_2^m (1-p_1-p_2)^{z-n-m}.
\end{align}$$
We deduce that
$$\begin{align}
P(N(A_1) = n, \, N(A_2) = m) &= \frac{p_1^n p_2^m e^{-\lambda}}{n!m!}\sum_{z=n+m}^\infty \frac{\lambda^z (1-p_1-p_2)^{z-n-m}}{(z-n-m)!} \\
&= \frac{p_1^n p_2^m \lambda^{n+m}e^{-\lambda}}{n!m!}\sum_{z=0}^\infty \frac{\lambda^z (1-p_1-p_2)^{z}}{z!} \\
&= \frac{p_1^n p_2^m \lambda^{n+m}}{n! m!} e^{-\lambda(p_1+p_2)} \tag{2}.
\end{align}$$
On the other hand, summing over all $m \in \mathbb{N}$ in $(2)$, we get that
$$P(N(A_1) = n) = \frac{(\lambda p_1)^n}{n!}e^{-\lambda p_1},$$
i.e. $N(A_1) \sim Po(\lambda p_1)$. Similarly $N(A_2) \sim Po(\lambda p_2)$. It follows that $$P(N(A_1) = n, N(A_2) = m) = P(N(A_1) =n)P(N(A_2) =m).$$
