Write your equation as
$$ s = 2 - \dfrac{2(2^p-1)}{n} + \dfrac{2^{p+1}-4}{n-1}$$
Since $\gcd(n,n-1) = 1$, we need $n$ to divide $2(2^p-1)$ and $n-1$ to divide $2^{p+1} -4$.
$n = 3$ works for any even $p$, with $s = (2^p+2)/3$.
$n=5$ works for $p$ divisible by $4$, with $s = (2^p + 14)/10$.
$n = 7$ works for $p \equiv 3 \mod 6$.
$n = 15$ works for $p \equiv 4 \mod 12$.
$n = 23$ works for $p \equiv 11 \mod 110$.
$n = 29$ works for $p \equiv 28 \mod 84$.
$n = 31$ works for $p \equiv 5 \mod 20$.
etc.