Three. (Full marks: 50 points)
Find all pairs of positive integers $(m, n)$ such that $m^{6}=n^{n+1}+n-1$
Three, Proof: We first prove a lemma.
Lemma: If $p$ is a prime of the form $6k-1$, then there does not exist an integer $a$ such that $p \mid (a^2 + 3)$.
Proof of the lemma: Suppose there exists an integer $a$ satisfying the condition, then
$$
(a+1)^3 + (a-1)^3 = 2a(a^2 + 3) \equiv 0 \pmod{p}
$$

Thus, $(a+1)^{6k} \equiv (a-1)^{6k} \pmod{p} \Rightarrow (a+1)^{p+1} \equiv (a-1)^{p+1} \pmod{p}$. By Fermat's Little Theorem, $(a+1)^2 \equiv (a-1)^2 \pmod{p} \Rightarrow a \equiv 0 \pmod{p}$. Since $p \mid (a^2 + 3)$, it follows that $p \mid 3$, hence $p = 3$; this contradicts the fact that $p$ is a prime of the form $6k-1$. The lemma is proved.

Returning to the original problem: First, we eliminate simple cases using common methods for solving indeterminate equations: When $n=1$, $m=1$. Now assume $n > 1$, then $m > 1$.
(1) When $2 \mid n+1$ or $3 \mid n+1$, if $2 \mid n+1$, let $n+1 = 2k$, then $\left(m^3 + n^k\right)\left(m^3 - n^k\right) = n - 1$. Since $n > 1$, we have $n - 1 = \left(m^3 + n^k\right)\left(m^3 - n^k\right) \geq m^3 + n^k > n$, which is a contradiction. Similarly, if $3 \mid n+1$, it also leads to a contradiction.
(2) When $3 \mid n$, then $m^6 \equiv -1 \pmod{3}$, which is a contradiction.
(3) When $n+1$ is a positive integer of the form $6k-1$, consider quadratic residues, then
$$
m^6 = (n+1-1)^{n+1} + (n+1) - 2 \equiv (-1)^{n+1} - 2 \equiv -3 \pmod{n+1}
$$

Thus, $(n+1) \mid (m^6 + 3)$. Since $n+1$ is a positive integer of the form $6k-1$, it must have at least one prime factor $p$ of the form $6k-1$ such that $p \mid (m^6 + 3)$. By the lemma, this situation is impossible, leading to a contradiction.
In summary, the positive integer solution to the original equation is $m = n = 1$.