The only natural numbers $n$ which satisfies the implication
$(1) \;\; d>0, d|n \;\;\; \Rightarrow \;\;\; d + 1 | n + 1$
are the odd primes.
Proof: Assume $d<n$ is a divisor of $n$. Then
$\frac{n}{d} - \frac{n+1}{d+1} = \frac{n - d}{d(d + 1)} \in \mathbb{N}$,
yielding
$\frac{n - d}{d(d + 1)} \geq 1$
which means
$(2) \;\; n > d^2$.
Let $p$ be the smallest prime divisor of $n$. By choosing ${\textstyle d = \frac{n}{p}}$ and inserting this in inequality (2) result in
$(3) \;\; {\textstyle \frac{n}{p} < p}$.
The fact that ${\textstyle \frac{n}{p}}$ is a divisor of $n$ and $p$ is the smallest prime divisor of $n$ combined with inequality (3) implies ${\textstyle \frac{n}{p}=1}$, i.e. $n=p$. Hence $d=1$ or $d=p$, which according to implication (1) give us $2 \, | \, p+1$ and $p+ 1 \,|\, p+1$ respectively. Hence $p$ is an odd prime. q.e.d.