Solution:

We will show it is divisible by $8$ and $3$. Since the least common multiple of $8$ and $3$ is $24$, this implies the result.

We factor $n^{2}-1 = (n-1)(n+1)$.

To show divisibility by $8$, note that $n-1$ and $n+1$ are two consecutive even integers. Among any two consecutive even integers, one of them must be divisible by $4$; the other one is divisible by $2$ by definition, so their product is divisible by $8$.

To show divisibility by $3$, note that $\{n-1, n, n+1\}$ form three consecutive integers. Thus at least one of them is divisible by $3$. We assumed $n$ was not divisible by $3$, so it must be either $n-1$ or $n+1$, hence their product is divisible by $3$ as well.