\[ \newcommand{\RR}{\mathbb{R}} \newcommand{\EE}{\mathbb{E}} \]
Suppose \(Z\sim \text{Poi}(\theta)\) and \(\theta\sim G\in \mathcal{P}(\mathbb{R}_+)\). The posterior mean of \(Z=z\) is given by: \[ \mathbb{E}[\theta \mid Z=z] = \frac{(z+1)f_G(z+1)}{f_G(z)}, \]
where \[ f_G(z) = \int e^{-\theta} \theta^{z}/z!dG(\theta). \]
We want to show that \(\mathbb{E}[\theta \mid Z=z]\) is monotonic in \(Z=z\). To do so, we want to show \[ \mathbb{E}[\theta \mid Z=z] \geq \mathbb{E}[\theta \mid Z=z-1], \]
which amounts to showing
\[ \begin{equation} \frac{\int e^{-\theta} \theta^{z+1}dG(\theta)}{\int e^{-\theta}\theta^{z}dG(\theta)} \geq \frac{\int e^{-\theta} \theta^{z}dG(\theta)}{\int e^{-\theta} \theta^{z-1}dG(\theta)}. \end{equation} \tag{1}\]
Idea: consider the measure \(dH(\theta)=e^{-\theta} \theta^{z-1} dG(\theta)\). By Cauchy-Schwarz, we have
\[ \left[ \int \theta \,dH(\theta)\right]^2 \leq \int \theta^2\, dH(\theta) \int 1\, dH(\theta), \]
which is equivalent to Equation 1.