Safe prime mod 24 Given a safe-prime $p = 2q + 1$ where $q$ is also a prime and $p \gt 7$, I've read in a crypto.se answer that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$.
I understand the proofs of why $p^2 \equiv 1 \pmod {24}$, and $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ for any prime $p$, and I can see that $p \equiv 11 \pmod {24}$ and $p \equiv 23 \pmod {24}$ are consistent with that, but can anyone explain why the other possible congruences for a prime $p$ (such as $p \equiv 1 \pmod {24}$) are excluded by $p$ being a safe prime?
My reasoning so far is:

For a prime $p \equiv 1 \pmod 6$ and  $p \equiv (1,7,13,19) \pmod {24}$, or $p \equiv 5 \pmod 6$ and $p \equiv (5,11,17,23) \pmod {24}$.
For a safe prime $p = 2q + 1$ it cannot be true that $p \equiv 1 \pmod 6$, otherwise $2q$ would be divisible by 6 and $q$ would not be prime. This eliminates 1, 7, 13 and 19.
Likewise $p = 2q+1 \equiv 5 \pmod {24}$ and $p = 2q+1 \equiv 17 \pmod {24}$ cannot hold, otherwise $q$ would have to be even: $q \equiv 2 \pmod {24}$ or $q\equiv 8\pmod {24}$ respectively.
This leaves $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ as possible congruences.

Is this correct and sufficient, and/or is there a better way of demonstrating that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ can and must hold?

A prime must be $\{1,5\} \text{ mod } 6$, so $\{1,5,7,11\} \text{ mod } 12$, so $\{1,5,7,11,13,17,19,23\} \text{ mod } 24$.  This is true for $q$.  Double each of these and add 1:
$1 \rightarrow 3 \not \in \{1,5,7,11,13,17,19,23\}$
$5 \rightarrow 11 \in \{1,5,7,11,13,17,19,23\}$
$7 \rightarrow 15 \not \in \{1,5,7,11,13,17,19,23\}$
$11 \rightarrow 23 \in \{1,5,7,11,13,17,19,23\}$
$13 \rightarrow 3 \not \in \{1,5,7,11,13,17,19,23\}$
$17 \rightarrow 11 \in \{1,5,7,11,13,17,19,23\}$
$19 \rightarrow 15 \not \in \{1,5,7,11,13,17,19,23\}$
$23 \rightarrow 23 \in \{1,5,7,11,13,17,19,23\}$
So the only options for $p$ are $11$ and $23 \text{ mod } 24$.
