The small primes $p\in \{3,5,7,13\}$ work, while $11 - \lfloor 11/7\rfloor 7 = 11 - 7 = 4 = 2^2$ doesn't. So assume $p>13$.
Reading the condition the other way, we need $p-4 = 3^n$, and $p-8, p-9$ divisible by just $2,5,7$; this forces $p-9 = 2^m$, or $p-8 = 7^m$ or $p-8 = 5^m$. The first cannot occur, since then $3^n = 2^m + 5$, impossible modulo $8$. The second cannot occur, since then $3^n = 7^m + 4$, impossible modulo $3$.
We are left with the case $3^n = 5^m + 4$, writing as $3^2(3^{n-2} - 1) = 5(5^{m-1} - 1)$. From $3^2 \mid 5^{m-1} - 1$ we need $6 \mid m-1$; but then $2^3\cdot 3^2\cdot 7\cdot 31 = 5^6 - 1 \mid 5^{m-1} - 1$, so we need $5\cdot 7 \mid 3^{n-2} - 1$; so $12\mid n-2$. But then $2^4\cdot 5\cdot 7\cdot 13\cdot 73 = 3^{12} - 1 \mid 3^{n-2} - 1$, so we need $73 \mid 5^{m-1} - 1$; but that forces $72\mid m-1$, and then $3^3 \mid 5^{72} - 1 \mid 5^{m-1} - 1$, so $3^3 \mid 3^2(3^{n-2} - 1)$, impossible. Therefore, no other solutions.