Can we say something about primes $p$ s.t. $p^p-2$ is prime?

a bit more detailed answer to,

I also had the idea that maybe if $p$ and $p^p-2$ are prime, maybe $$(p^p-2)^{p^p-2}-2$$ is prime again?

Let $p$ an odd prime s.t. $p^p-2$ is also prime, so $p\equiv -1 , 1 \pmod{3}$

If $p \equiv -1 \pmod{3}$, then $p^p \equiv (-1)^p \equiv -1 \pmod{3} $ (since $p$ is odd)

so $p^p-2 \equiv -1-2 \equiv 0 \pmod 3$ , but $p^p-2$ is an odd prime $\ge 3^3-2>3$

thus $p \equiv 1 \pmod{3}$, However $q=7^7-2=823541$ is prime

but $q\equiv -1 \pmod{3}$ so $q^q-2$ isn't prime anymore , hence $(7^7-2)^{7^7-2}-2$ is not prime !

In general, we can do similar elementary restrictions on odd $p$ by noticing that :

$$ 5 \mid p^p-2 \Leftrightarrow p \equiv 3,17 \pmod{20} $$

$$ 7 \mid p^p-2 \Leftrightarrow p \equiv 11,16 \pmod{21} $$

$$ 11 \mid p^p-2 \Leftrightarrow p \equiv 39,73,101,107 \pmod{110} $$

$$ 13 \mid p^p-2 \Leftrightarrow p \equiv 59,115,145,149 \pmod{156} $$

$$ 17 \mid p^p-2 \Leftrightarrow p \equiv 59,111,117,121 \pmod{136} $$

This can speed up research a little bit , but the main question right now is :

Can you find a prime $p>19$ s.t. $p^p-2$ is prime ?

However, we can pick $p>2500$ as Robert Price stated in A100408

Note that $2500$ research zone on $p$ is still very small , and so another small candidate $p$ might appear below $10^6$

but $2500^{2500} \approx 7.079811261 \times 10^{8494}$ , and so testing primality isn't that easy anymore 💀

I'd suspect only finite candidates to exist as @Dean Menezes explained heuristically :-)