Let $ n, m$ be positive integers of different parity, and $ n > m$. Find all integers $ x$ such that $ \frac {x^{2^n} - 1}{x^{2^m} - 1}$ is a perfect square.
The given expression reduces to $ (x^{2^m} +1)(x^{2^{m+1}}+1)...............(x^{2^{n-1}} +1)$
We see clearly that there are an odd number of terms in the expansion because m and n have different parity.
So the expression is divisible by some $ 2^l$ where l is odd if x is odd
because each term of the expression can be written as $ 2y$ where y is odd
So the given expression can never be a perfect square if x is odd.
If x is even observe that each term in the expression is odd and each term is relatively prime to each other.
So then each term should be a perfect square
But $ (x^{2^m}$ is itself a perfect square so if $ (x^{2^m} +1)$ is a perfect square then x is 0.
I guess this is the only answer.