First of all, $n$ should be an odd prime. If $n+1$ has an odd prime divisor $p$ then for $k=\frac{p+1}{2}<p\leq \frac{n+1}{2}<n$ the number $4k^2+n$ is divisible by $p$ and more than $p$. So $n+1=2^s$ for some positive integer $s$. It is easy to check that $n=3,7$ are charming and $n=1$ is not. Let us consider that $s\geq 4$ and $n\geq 15$. Then $n+9=2^s+8$ is not a power of $2$, so it has some odd prime divisor $q$. Let us take $2<k=\frac{q+3}{2}\leq q\leq \frac{n+9}{2}<n$, then $4k^2+n$ is divisible by $q$ and more than $n+9>q$.