We claim that if $n \geq 10$, then $2 n+3 \nmid 2^{n!}-1$ if and only if both $n+1$ and $2 n+3$ are prime. If both $n+1$ and $2 n+3$ are prime, then assume $2 n+3 \mid 2^{n!}-1$. By Fermat Little Theorem, $2 n+3 \mid 2^{2 n+2}+1$. However, since $n+1$ is prime, $\operatorname{gcd}(2 n+2, n!)=2$, so $2 n+3 \mid 2^{2}-1=3$, a contradiction. If $2 n+3$ is composite, then $\varphi(2 n+3)$ is even and is at most $2 n$, so $\varphi(2 n+3) \mid n$ !, done. If $n+1$ is composite but $2 n+3$ is prime, then $2 n+2 \mid n$ !, so $2 n+3 \mid 2^{n!}-1$. The prime numbers between 50 and 100 are $53,59,61,67,71,73,79,83,89,97$. If one of these is $n+1$, then the only numbers that make $2 n+3$ prime are 53,83 , and 89 , making $n$ one of 52,82 , and 88 . These sum to 222.