= Solution
Writing $y_i$ for the observed counts, independence of the <Poisson distributions> gives the <log-likelihood>
$$
\ell(\beta)=\sum_{i=1}^n\left\{y_i\beta x_i-e^{\beta x_i}-\log(y_i!)\right\}.
$$
Its <score function> and negative second derivative are
$$
U(\beta)=\sum_ix_i(y_i-e^{\beta x_i}),\qquad J(\beta)=\sum_ix_i^2e^{\beta x_i}.
$$
Unless all $x_i$ vanish, $J(\beta)>0$, so the <log-likelihood> is strictly concave. Use <Newton method>, starting from a finite $\beta_0$, with
$$
\boxed{\beta_{r+1}=\beta_r+\frac{\sum_ix_i(y_i-e^{\beta_rx_i})}{\sum_ix_i^2e^{\beta_rx_i}}.}
$$
Stop when the <score function> or the change in $\beta$ is small; a reduced step can ensure that the <log-likelihood> increases. Any finite zero of the <score function> is the unique <maximum-likelihood estimator>.
Existence needs a qualification. If the nonzero $x_i$ include both signs, $U$ decreases from $+\infty$ to $-\infty$, so a finite root exists. If all nonzero $x_i$ are positive, a finite root exists exactly when $\sum_ix_iy_i>0$; otherwise the supremum is approached as $\beta\to-\infty$. The corresponding condition for all negative $x_i$ is $\sum_ix_iy_i<0$, with a supremum at $+\infty$ otherwise. When all $x_i=0$, the parameter is unidentifiable. These cases explain when the iterative algorithm actually has a finite estimate to find.
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