= Solution
Conditioning on $Z=z$ has positive <probability> exactly when $z=2r+1$ for an <integer> $r\geq0$. This is the same <event> as $X+Y=r$. The <addition of independent Poisson random variables> gives a <Poisson distribution> with parameter $\lambda+\mu$ for the sum. Therefore, for $0\leq t\leq r$,
$$
\begin{aligned}
\mathbb P(X=t\mid Z=2r+1)
&=\frac{e^{-(\lambda+\mu)}\lambda^t\mu^{r-t}/[t!(r-t)!]}
{e^{-(\lambda+\mu)}(\lambda+\mu)^r/r!}\\
&=\binom rt\left(\frac{\lambda}{\lambda+\mu}\right)^t
\left(\frac{\mu}{\lambda+\mu}\right)^{r-t}.
\end{aligned}
$$
Thus \b[$X\mid\{Z=2r+1\}\sim\operatorname{Binomial}(r,\lambda/(\lambda+\mu))$]. The case $r=0$ is a constant zero <random variable>. For even, nonpositive, or noninteger $z$, the conditioning <event> has zero <probability>, so an event-conditional <probability distribution> is not specified by the quotient formula. This is a two-component case of <Poisson-multinomial conditioning>.
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