HINT $\rm\: \mod\ 24\::\ \ x^2 \equiv 1\ \Rightarrow\ (5x)^2 \equiv 1\:,\ $ but $\rm\:5\:x\:$ is prime iff $\rm\: x= \pm1$
Note that this yields a general structural reason explaining why such integers can't all be primes. Namely, the integers you describe are simply those integers that, when reduced modulo $24\:,$ yield square roots of $1\:.\:$ But such roots are closed under multiplication: $\rm\ x^2\equiv 1,\ y^2\equiv 1\ \Rightarrow\ (xy)^2\equiv 1\:.\:$ But primes are not closed under multiplication. For example, one can take any of your prime solutions and multiply them to obtain a composite solution, e.g. $\rm\ 5^2 = 25,\: \ 5\cdot 7 = 35\:,\:$ etc.
Notice that there are precisely $8\:$ square-roots of $\rm 1\ (mod\ 24)\ $ viz. $\rm \pm 1,\:\pm 5,\:\pm 7,\: \pm 11\:,\:$ corresponding (by $\rm CRT$) to the product of the two roots $\rm\ \pm 1\ (mod\ 3)\ $ times the four roots $\rm\ \pm 1,\: \pm 3\ (mod\ 8)\:.\:$ Note that these are precisely the congruence classes of all the integers coprime to $\:3\:$ and $\rm\:2\:,\:$ which includes all primes $> 3$. This explains your empirical observations above. The key observation, that $\rm\ x^2\equiv 1\ (mod\ 24)\ \iff\ x\:$ is coprime to $\:6\:,\:$ is nothing but a very special case computation of Carmichael's generalization of Euler's phi-function - see my post here for details.