= Solution
Use the ideal detection convention of one collected <electron> per detected <photon>; otherwise replace <photon> totals by detected <electron> totals using the <quantum efficiency>. Let $A$ be the summed measurement on the <star> patch and $B'$ the independent background-only patch. After additive offsets are removed, their means are $Q+B$ and $B$. Independent <Poisson distribution> counts have <variance> equal to their mean, and a single read of $n$ <detector pixels> contributes $nr^2$ to each patch's <variance>. Therefore
$$
\mathbb E(A-B')=Q,\qquad\operatorname{Var}(A-B')=Q+2B+2nr^2,
\qquad
\boxed{Z=\frac{Q}{\sqrt{Q+2B+2nr^2}}.}
$$
The two factors of two arise from the independently measured background and the second patch's <read noise>, not from doubling the stellar signal. The formula assumes equal background means, independent <detector pixels> and patch measurements, one read per patch, and negligible <dark current> or other noise. Several exposure reads or correlated <detector pixels> require their own <variance> and <covariance> terms.
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