(Note that in the following I am quite careful about when to use capital $\Lambda$ and when to use lower-case $\lambda$, and when to use capital $X$ and when to use lower-case $x$.)
Suppose the distribution of $\Lambda$ is the Gamma distribution:
$$
\frac 1 {\Gamma(\alpha)} \left( \frac \lambda \beta \right)^{\alpha-1} e^{-\lambda/\beta} \left(\frac{d\lambda}\beta\right).
$$
And suppose the conditional distribution of $X$ given $\Lambda$ is the Poisson distribution
$$
x \mapsto \frac{e^{-\Lambda} \Lambda^x }{x!} \text{ for } x = 0,1,2,3,\ldots
$$
Then you have
\begin{align}
& \Pr(X=x) \\[10pt]
= {} & \operatorname{E}(\Pr(X=x \mid \Lambda)) = \operatorname{E}\left( \frac{e^{-\Lambda} \Lambda^x}{x!} \right) \\[10pt]
= {} & \int_0^\infty \left( \frac{e^{-\lambda}\lambda^x}{x!} \right) \frac 1 {\Gamma(\alpha)} \left( \frac \lambda \beta \right)^{\alpha-1} e^{-\lambda/\beta} \left(\frac{d\lambda}\beta\right) \\[10pt]
= {} & \frac 1 {x!\Gamma(\alpha)\beta^\alpha} \int_0^\infty \lambda^{x+\alpha-1} e^{-\lambda(1\,+\,1/\beta)} \, d\lambda \\
& \qquad \qquad \text{(We pulled out the factors that do not depend on $\lambda$).}\\[10pt]
= {} & \frac 1 {x!\Gamma(\alpha)\beta^\alpha} \cdot \left(\frac \beta {\beta+1} \right)^{x+\alpha} \int_0^\infty \left( \lambda \left( 1 + \frac 1 \beta \right) \right)^{x+\alpha-1} e^{-\lambda(1+1/\beta)} \left( d\lambda \left( 1 + \frac 1 \beta \right) \right) \\[10pt]
= {} & \frac 1 {x!\Gamma(\alpha)\beta^\alpha} \cdot \left(\frac \beta {\beta+1} \right)^{x+\alpha} \int_0^\infty u^{x+\alpha-1} e^{-u} \, du \\[10pt]
= {} & \frac {\Gamma(x+\alpha)} {x!\Gamma(\alpha)\beta^\alpha} \cdot \left(\frac \beta {\beta+1} \right)^{x+\alpha} \\[10pt]
= {} & \frac {\Gamma(x+\alpha)} {x!\Gamma(\alpha)} \cdot \left( \frac 1 {\beta+1} \right)^\alpha \left(\frac \beta {\beta+1} \right)^x
\end{align}
and this is a negative binomial probability mass function with $p$ and $q$ equal to $\dfrac 1 {\beta+1}$ and $\dfrac\beta{\beta+1}$, not necessarily respectively (depending on which convention you follow). This is the probability that the number of failures before the $\alpha$th success is $x$, when the probability of success on each independent trial is $1/(\beta+1).$
So a Gamma mixture of Poisson distributions is a negative binomial distribution, and that is the connection between Poisson and negative binomial.
