There most certainly is a relation between the two!
In general, a stochastic process is simply an indexed collection of random variables defined on a probability space $(\Omega, \mathcal F,\mathbb P)$, e.g. $N=\{N(t):t\in I\}$ for some index set $I$. A Poisson process is one of the most well-known examples of an arrival process, which satisfies the following properties (with probability $1$, or "almost surely"):
- $N(0) = 0$
- If $s<t$, then $N(s)\leqslant N(t)$.
- The map $t\mapsto N(t)$ is right-continuous.
- If $\lim_{t\to t_0^-} N(t)<N(t_0)$, then $N(t_0)=N(t)+1$.
The fourth condition describes what are called jumps, that is, the random times at which arrivals occur. If we set $T_0=0$ and
$$T_{n+1} = \inf\{t>T_n : N(t)>N(T_n)\} $$
for $n=0,1,2,\ldots$ then $T_n$ are the arrival times of the process $N(t)$. The times between arrivals $T_{n+1}-T_n$ are called interarrival times. A Poisson process is an arrival process which also satisfies the following conditions:
- For any $s,t\geqslant 0$ and $0\leqslant u_1<\cdots<u_n\leqslant t$, $N(t+s)-N(t)$ is independent of $\{N(u_1), N(u_2),\ldots, N(u_n)\}$.
- For any $s,t\geqslant 0$, $N(s+t)-N(t)$ is independent of $t$.
These are referred to as independent increments and stationary increments, respectively. From these conditions (and assuming that $N(t)$ is not identically zero) it can be shown that there exists $\lambda>0$ such that $\mathbb P(N(t)=k) = \frac{e^{-\lambda t}(\lambda t)^k}{k!}$ for all $t>0$, $k=0,1,2,\ldots$ (for brevity, I will omit the argument, which can be found in any text on stochastic processes). This $\lambda$ is the rate or intensity of the Poisson process.
From there, we can see that for $s,t> 0$ and $k=0,1,2,\ldots$,
$$\mathbb P(N(s+t)-N(t)=k) = \frac{e^{-\lambda s}(\lambda s)^k}{k!}, $$
that is, the distribution of arrivals in a time interval of length $s$ follows a Poisson distribution with rate $\lambda s$. To derive the interarrival distribution, note that for for any $n\geqslant 0$ and $t>0$,
$$\mathbb P(T_{n+1}-T_n>t) = \mathbb P(T_1>t) $$
due to stationary increments, and
$$\mathbb P(T_1>t) = \mathbb P(N(t) = 0) = e^{-\lambda t}, $$
hence the time between arrivals is exponentially distributed with rate $\lambda$.
