= Solution
Let $N_j(A)$ be the number of particles of gas $j$ in a bounded <Lebesgue measurable set> $A$. Because Lebesgue measure is diffuse, two independent spatial <Poisson point process>[Poisson point processes] have no common particle almost surely. Hence the mixture count is
$$
N(A)=N_1(A)+N_2(A).
$$
Writing $\lambda_j=z_j|A|$, independence gives the <probability generating function>
$$
\begin{aligned}
\mathbb E[s^{N(A)}]
&=\mathbb E[s^{N_1(A)}]\mathbb E[s^{N_2(A)}]\\
&=\exp(\lambda_1(s-1))\exp(\lambda_2(s-1))\\
&=\exp\bigl((z_1+z_2)|A|(s-1)\bigr).
\end{aligned}
$$
Thus
$$
N(A)\sim\operatorname{Pois}\bigl((z_1+z_2)|A|\bigr).
$$
If $A_1,\ldots,A_m$ are pairwise disjoint, the family
$$
\{N_j(A_r):j=1,2,\ 1\leq r\leq m\}
$$
is independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums $N(A_r)$ are independent. These two facts prove directly that the mixture is a Poisson point process. By the <Superposition theorem for Poisson point processes>, its activity is
$$
\boxed{z_1+z_2}.
$$
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