= Solution
Let $T_i=X_i+Y_i$. Since the two counts are independent <Poisson distributions> with means $\lambda_i$ and $\beta\lambda_i$, their total is a <Poisson distribution> with mean $(1+\beta)\lambda_i$. For a fixed observed total $t_i=x_i+y_i$, divide the joint <probability mass function> by that of the total:
$$
\Pr(Y_i=y_i\mid T_i=t_i)
=\frac{e^{-(1+\beta)\lambda_i}\lambda_i^{x_i}(\beta\lambda_i)^{y_i}/(x_i!y_i!)}
{e^{-(1+\beta)\lambda_i}[(1+\beta)\lambda_i]^{t_i}/t_i!}
=\binom{t_i}{y_i}\frac{\beta^{y_i}}{(1+\beta)^{t_i}}.
$$
Define $p=\beta/(1+\beta)$, so $1-p=1/(1+\beta)$. Then
$$
Y_i\mid T_i=t_i\sim\operatorname{Bin}(t_i,p).
$$
For independent patients, the <paired Poisson conditional likelihood> is
$$
\boxed{L_c(p)=\prod_i\binom{t_i}{y_i}p^{y_i}(1-p)^{t_i-y_i}
\ \propto\ p^{\sum_i y_i}(1-p)^{\sum_i x_i}.}
$$
Every nuisance rate $\lambda_i$ cancels, while each patient's own baseline count level is retained through the conditioning total. A patient with total zero would contribute the constant one.
For these data, $\sum_i x_i=161$, $\sum_i y_i=23$, and $\sum_i t_i=184$. Maximizing this <conditional likelihood> gives
$$
\widehat p=\frac{23}{184}=\frac18,\qquad
\widehat\beta=\frac{\widehat p}{1-\widehat p}
=\frac{23}{161}=\frac17.
$$
These are the fit under the model without a cure component, not the estimates of the mixture model in the later parts.
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