We prove that the numbers that have the given property are $1$ and the odd prime numbers. It is clear that all these numbers do indeed have the desired property and also that $2$ does not have it.

Conversely, let us consider a composite number $n$ and prove that it does not have the given property. If $n$ is composite, then $n = ab$ such that $1 < a \le b < n$. It follows that $b + 1$ divides $n + 1$, i.e. there exists $c \in \mathbb{Z}$ such that $c(b + 1) = n + 1 = ab + 1$. We obtain that $b$ divides $c - 1$. Obviously, $c > 1$. We deduce that $c - 1 \ge b$, i.e. $c \ge b + 1$. Then $ab + 1 = c(b + 1) \ge (b + 1)^2$, which means that $ab + 1 \ge b^2 + 2b + 1$, leading to $a \ge b + 2$, which contradicts $a \le b$.

In conclusion, no composite number does satisfy the requirements.