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It may be worth saying something about generic $p$. For such primes $p-1$ will have many prime factors, which implies that $(\mathbf{Z}/(p-1)\mathbf{Z})^\times$ will surject onto $(\mathbf{Z}/2\mathbf{Z})^m$ for some large $m$. Thus not many elements of $(\mathbf{Z}/(p-1)\mathbf{Z})^\times$ are of the form $y^2$, so not many elements $x$ of $\mathbf{Z}/p\mathbf{Z}$ are even of the form $2^{y^2}$, let alone of the form $2^{2^{2^z}}$ for some integer $z \equiv x\pmod{p}$.