= Solution
Given $\Theta=\theta$, the sum of the independent yearly counts is $S_n\sim\operatorname{Poisson}(n\theta)$: their <probability generating functions> multiply to $\exp(n\theta(z-1))$. Averaging this conditional law over the <prior> gives the <mixed Poisson distribution>
$$
\boxed{p_n(s):=\mathbb P(S_n=s)
=\frac{n^s}{s!}\int_0^\infty e^{-n\theta}\theta^s\pi(\theta)\,d\theta,\qquad s\geq0.}
$$
The likelihood for the full vector depends on $\theta$ only through $s$, so $S_n$ is a <sufficient statistic>; conditioning on the total or on the full observed vector gives the same <posterior> for $\Theta$.
For an event of positive <probability> $p_n(s)>0$, the <posterior density> is proportional to $e^{-n\theta}\theta^s\pi(\theta)$. Put $I_s=\int_0^\infty e^{-n\theta}\theta^s\pi(\theta)\,d\theta$. Its mean is $I_{s+1}/I_s$. The future count has <conditional mean> $\Theta$, so
$$
\begin{aligned}
\mathbb E[X_{n+1}\mid S_n=s]
&=\frac{I_{s+1}}{I_s}\\
&=\frac{(s+1)!\,p_n(s+1)/n^{s+1}}{s!\,p_n(s)/n^s}
=\boxed{\frac{(s+1)\mathbb P(S_n=s+1)}{n\mathbb P(S_n=s)}}.
\end{aligned}
$$
This is the <posterior mean from adjacent mixed Poisson probabilities>. Both <probabilities> refer to the same $n$-year exposure; the numerator is not a <probability> for $S_{n+1}$. For a proper <prior> and $n>0$, the integrals defining this <posterior mean> are finite because $\theta^{s+1}e^{-n\theta}$ is bounded on $[0,\infty)$.
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