From my thinking so far, there is no solution. Is this an open question or is the answer well known?
Here's my reasoning about this issue:
If a solution exists, then:
$$p^c(p^{a-c} - 1) = q^d(q^{b-d} - 1)$$
Since $p^c \mid (q^{b-d}-1)$ and $q^d \mid (p^{a-c}-1)$, it follows that:
$$q^{b-d} \equiv 1 \pmod {p^c}$$
and
$$p^{a-c} \equiv 1 \pmod {q^d}$$
Assuming that we want to minimize $b-d$ and $a-c$, using Carmichael's Theorem, it follows that:
$$b-d = \varphi(p^c) = (p-1)p^{c-1}$$
$$a-c = \varphi(q^d) = (q-1)q^{d-1}$$
Now, this looks preposterous to me:
$$p^c - q^d = p^{c+(q-1)q^{d-1}} - q^{d+(p-1)p^{c-1}}$$
where $c,d \ge 2$ and $p,q \ge 3$
If we assume a multiple of $\varphi(p^c)$ and $\varphi(q^d)$, it still looks wrong to me:
$$p^c - q^d = p^{c+u(q-1)q^{d-1}} - q^{d+v(p-1)p^{c-1}}$$
where $c,d \ge 2$ and $p,q \ge 3$ and $u,v \ge 1$
Is there a straight forward way to show this is an impossible equation where $p,q,c,d$ are all positive integers? Is this a tougher problem than it seems?
Thanks very much,
-Larry
Edit: I added additional information to make it clearer why $b-d = \varphi(p^c)$ David is exactly right. If a solution exists, it will be a multiple.
I believe that my question is addressed by Michael A. Bennett in his paper On Some Exponential Equations of S.S. Pillai
As I understand it, the answer may be yes. Bennett shows that for $a^x - b^y = c$ where $a,b,c$ are nonzero integers with $a,b \ge 2$, then the equation has at most $2$ solutions in positive integers $x$ and $y$.