Solution:

a.
The required pairs are: $\left(3, 3^{667}\right)$, $\left(23, 23^{87}\right)$, $\left(29, 29^{69}\right)$.

Indeed, let $x = p$ be a prime number. Then from the equation it follows that in the factorization of $y$ there cannot be prime factors different from $p$, hence $y = p^{n}$ for some $n$. Substituting into the equation we thus have $p^{2001} = p^{n p}$, from which $2001 = n p$. It follows that $p$ must be a prime factor of $2001$ and $n = 2001 / p$. In this way we obtain the three pairs listed.

b.
The required pairs are
$(1, 1), \left(3, 3^{667}\right), \left(23, 23^{87}\right), \left(29, 29^{69}\right), \left(69, 69^{29}\right), \left(87, 87^{23}\right), \left(667, 667^{3}\right), (2001, 2001)$.

Let $p$ be a prime that appears with exponent $n$ in the factorization of $x$. Then from the equation we have that $p$ also appears in the factorization of $y$, with an exponent that we denote by $k$. We can therefore write $x = p^{n} a$, $y = p^{k} b$, where $a$ and $b$ are not divisible by $p$. Substituting into the equation we find that $p^{2001 n} \cdot a^{2001} = p^{k p^{n} a} \cdot b^{p^{n} a}$. Equating the exponents relative to the prime $p$, we thus have that $2001 n = k a p^{n}$, and hence $p^{n}$ divides $2001 n$. We now have two cases.

- If $p$ does not divide $2001$, then $p^{n}$ must divide $n$, which is impossible since $p^{n} > n$ (since $p^{n} \geq 2^{n} = (1+1)^{n} = 1^{n} + n \cdot 1^{n-1} + \ldots > n$).
- If $p$ is a factor of $2001$, then $p^{n-1}$ must divide $n$, which is possible only for $n = 1$. Indeed for $n > 1$ we have $p^{n-1} \geq 2^{n-1} = (1+1)^{n-1} = 1^{n} + (n-1) \cdot 1^{n-1} + \ldots > n$.

We have thus shown that $p$ can only be $3$, $23$, or $29$, and must appear with exponent $1$ in the factorization of $x$. Therefore $x$ is necessarily a divisor of $2001$.

On the other hand if $d$ is any divisor of $2001$, setting $e = 2001 / d$ one immediately obtains $d^{2001} = d^{e d} = \left(d^{e}\right)^{d}$, so the pair $\left(d, d^{e}\right)$ is a solution of the equation.

Second solution

Let $d$ be the G.C.D. of $x$ and $2001$, and set $x = d a$, $2001 = d e$, so that $a, e$ are coprime. The equation simplifies to $x^{e} = y^{a}$, and hence we have $x = z^{a}$, for some suitable $z$, as can be seen by comparing the exponents of each prime factor of $x$ on the two sides of the equation.

Since $x$ and $z$ have the same prime factors and $d$, being a divisor of $2001$, is a product of distinct primes, we also have $z = d b$, and hence $d a = (d b)^{a}$, that is $a = b^{a} d^{a-1}$.

This implies that $x = d$, and hence $x$ divides $2001$. Indeed if $b > 1$, then $b^{a} > a$, while if $b = 1$ and $a, d > 1$, then $d^{a-1} \geq 3^{a-1} > 2^{a-1} \geq a$.

(Recall that for every $b \geq 2$ and every integer $a \geq 0$ we have $b^{a} = (1 + (b-1))^{a} = 1 + a(b-1) + \ldots \geq a + 1$).

On the other hand if $d$ is any divisor of $2001$, setting $e = 2001 / d$ one immediately obtains $d^{2001} = d^{e d} = \left(d^{e}\right)^{d}$, so the pair $\left(d, d^{e}\right)$ is a solution of the equation.