Here's what I have so far

Without loss of generality, assume that P < Q.

P and Q must both be powers of a common factor, x.

Let P = x^k, Q = x^m, k < m

Substituting gives

(x^k)^(x^k +2x^m) = (x^m)^(x^m + 2x^k)

Associately,

x^(kx^k +2kx^m) = x^(mx^m + 2mx^k)

So kx^k +2kx^m = mx^m + 2mx^k

Rearranging gives

x^(m-k) = (2m-k)/(2k-m)

Because the lhs and the numerator are positive, it follows that

the denominator (2k - m) is also positive, so

k < m < 2k

But that's as far as I got.

Can anybody solve

x^(m-k) = (2m-k)/(2k-m)

where x, m and k are positive integers and k < m < 2k?

Or how about letting a = m - k?

Then

x^a = (2a + k)/(k - a)

where x, a and k are positive integers and a < k ?