 Solution

Ignoring terms independent of $\theta$, the log-likelihood isIts score function isSince ${\ell }^{\prime \prime }(\theta )=-({\sum }_i{Y}_i)/{\theta }^2\le 0$, the interior critical point is the maximum, givingThe sum of independent Poisson distributions is Poisson with mean $\theta {\sum }_i{x}_i$, soThus the maximum-likelihood estimator is also unbiased.