= Solution
Given the parameters and $Z$, the coordinates are conditionally independent. Write $m_i=x_i^T\beta+\sigma^2Z_i$, $v=\sigma_0^2$, and let $y_i$ denote the observed count. Their full <conditional distributions> have scalar densities
$$
\boxed{q_i(t)\propto\exp\left[y_it-n_i\log(1+e^t)-\frac{(t-m_i)^2}{2v}\right],\qquad t\in\mathbb R.}
$$
Each is proper: the <binomial distribution> likelihood factor is bounded, and the remaining factor is a proper <normal distribution> density.
One completely specified exact algorithm is <rejection sampling> from $N(m_i,v)$. Define
$$
H_i(t)=\frac{e^{y_it}}{(1+e^t)^{n_i}},\qquad
M_i=\sup_tH_i(t).
$$
For $0<y_i<n_i$,
$$
M_i=\left(\frac{y_i}{n_i}\right)^{y_i}
\left(1-\frac{y_i}{n_i}\right)^{n_i-y_i};
$$
for $y_i=0$ or $y_i=n_i$, take $M_i=1$, and for $n_i=0$ take $H_i=M_i=1$. Propose $t\sim N(m_i,v)$ and accept with probability $H_i(t)/M_i$, repeating until acceptance. The accepted density is proportional to the proposal times $H_i$, exactly $q_i$. Its acceptance probability is positive, so it terminates almost surely, although it can be inefficient.
An efficient exact alternative is <adaptive rejection sampling>. The log density has
$$
\ell_i'(t)=y_i-n_i\frac{e^t}{1+e^t}-\frac{t-m_i}{v},\qquad
\ell_i''(t)=-n_i\frac{e^t}{(1+e^t)^2}-\frac1v<0.
$$
Thus it is a <log-concave probability density>. Choose one tangent point with positive derivative in the left tail and one with negative derivative in the right tail; these exist because the Gaussian term dominates there. Their tangent upper hull gives an integrable piecewise exponential rejection envelope. Sample that envelope, accept using the target-to-envelope ratio, and add evaluated points to improve it. Concavity guarantees the upper bound, so accepted samples remain exact. Sample each coordinate independently by either method; a finite number of Metropolis steps would not provide the requested exact draws.
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