Given odd primes $p\ne q$, by the CRT we can find an integer $x$ such that $x\equiv 2^{p-1}\pmod q$ and $x\equiv 2^{q-1}\pmod p$. Can this procedure be reversed?

For which integers $x$ there exist odd primes $p\ne q$ satisfying $$ \begin{cases} 2^{p-1}\equiv x\pmod q \\ 2^{q-1}\equiv x\pmod p \end{cases}\ ? $$

If such $p$ and $q$ exist, then I call $x$ *nice*; otherwise, $x$ is *tough*. Say, $1$ is nice (as witnessed by the prime pair $(11,31)$), while $0$ is tough.

Computations show that in the range $-20\le x\le 20$, the numbers $$ -19, -18, -17, -16, -13, -11, -8, -6, -5, -4, 1, 2, 3, 4, 6, 8, 9, 10, 13, 14, 16, 17, 18, 19 $$ are nice, while $$ -20, -15, -14, -12, -10, -9, -7, -3, -2, -1, 0, 5, 7, 11, 12, 15, 20 $$ are good candidates to be tough, in the sense that there are no primes $3\le p<q<10^4$ such that the congruences above hold true.

This suggests a number of questions.

Is there an algorithm to determine whether a given integer $x$ is nice or tough?

Do positive tough integers exist? If so, what is the smallest of them? (It is quite likely that $5$ is tough, but I couldn't prove this.)

Are there infinitely many tough integers of both signs? Are there "more" negative than positive tough integers, as computations suggest?

I do not ask whether negative tough integers exist since I know that $x=-1$ is tough; here is the proof.

Let $d$ denote the order of $2$ in $(\mathbb Z/q\mathbb Z)^\times$. If $2^{p-1}\equiv -1\pmod q$, then $d\nmid p-1$, but $d\mid 2(p-1)$; consequently, $\nu_2(d)>\nu_2(p-1)$ where $\nu_2$ is the $2$-adic valuation function. On the other hand, from $d\mid(q-1)$ we get $\nu_2(d)\le \nu_2(q-1)$. Hence, $\nu_2(p-1)<\nu_2(q-1)$. Switching the roles of $p$ and $q$, in the very same way we obtain $\nu_2(q-1)<\nu_2(p-1)$, a contradiction.

My motivation came originally from this problem (consider $n=pq$), but it will be fair to say that I am driven mainly by curiosity.