Let's say $n=pk$ for the smallest divisible prime $p$ This means that $k+1|pk+1\implies gcd(k+1,pk+1)=k+1$. By Euclidean algorithm and other gcd properties, our gcd simplifies to $$gcd(k+1,(p-1)k)$$$$gcd(k+1,p-1)$$If $n$ is not a prime, then clearly $k\geq p$ and hence $gcd(k+1,p-1)\leq p-1<k+1$. Hence, $n$ cannot be composite. We can easily see that all odd primes work.

If $n$ is composite, then $n=ab$ where $a,b>1$. The condition gives $$ab+1 \equiv (a-(a+1))b + 1\equiv 1-b \equiv b-1 \equiv 0 \pmod{a+1} \implies a+1|b-1 \implies b-1\ge a+1 \implies b-a \ge 2$$Similarly, $a-b \ge 2$, a contradiction. So only primes.. Another way is $a+1|ab+1-b(a+1)=1-b \implies |1-b| \ge |a+1| \implies b-1 \ge a+1$.