= Solution
Conditional on $\Theta=\theta$, summing the individual independent Poisson counts gives $Y_j\sim\operatorname{Poisson}(m_j\theta)$. The between-year <conditional independence> makes the <likelihood function>
$$
L(\theta)=\prod_{j=1}^n\frac{e^{-m_j\theta}(m_j\theta)^{y_j}}{y_j!}
\ \propto\ \theta^y e^{-W\theta},\qquad
y=\sum_jy_j,\quad W=\sum_jm_j.
$$
Multiplying by the <gamma distribution> prior density, proportional to $\theta^{\alpha-1}e^{-\beta\theta}$, gives the <Poisson-gamma conjugacy with unequal exposures>:
$$
\boxed{\Theta\mid y_1,\ldots,y_n
\sim\operatorname{Gamma}(\alpha+y,\beta+W).}
$$
For an action $t$, the posterior <squared-error loss> decomposes as
$$
\mathbb E[(\Theta-t)^2\mid\mathbf y]
=\operatorname{Var}(\Theta\mid\mathbf y)
+\bigl(t-\mathbb E[\Theta\mid\mathbf y]\bigr)^2.
$$
Hence the <Bayes estimator under squared error loss> is the <posterior mean>, giving
$$
\boxed{\widehat\theta_{\rm Bayes}=\frac{\alpha+y}{\beta+W}.}
$$
Future counts are independent of the observed years conditional on $\Theta$, so the <law of total expectation> then gives the posterior predictive expected count
$$
\boxed{\mathbb E[Y_{n+1}\mid\mathbf y]
=m_{n+1}\mathbb E[\Theta\mid\mathbf y]
=m_{n+1}\frac{\alpha+\sum_jy_j}{\beta+\sum_jm_j}.}
$$
\b[The Bayesian and credibility estimates coincide exactly.] This <exact Bühlmann–Straub credibility for Poisson-gamma counts> occurs because the posterior mean is already affine in the exposure-weighted experience, and therefore belongs to the class over which the credibility estimate minimizes <mean squared error>.
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