Exponential Divisibility Problems Find all positive integers $m$ and $n$ such that $mn$ divides $(2^{2^m}+1)(2^{2^n}+1)$
$$$$ Let $p$ be any arbitrary prime factor of $m$, then $p$ divides any one of $2^{2^m}+1$ or $2^{2^n}+1$. Suppose $p$ divides $2^{2^m}+1$. Now let $x$ be the smallest positive integer such that $$2^x \equiv -1modp$$ and $y$ be the smallest positive integer such that $$2^y \equiv 1 modp$$. Now it is easy to see that both $x$ and $y$ exists and $y > x$. Also $2^m \geq x$. Now if $2^m>x$ then let $$2^m=kx+l$$, $0 \leq l<x$ So we get $$-1 \equiv 2^{2^m}=2^{kx+l} \equiv (-1)^k2^l modp$$ Now if $k$ is even then we get $$2^l \equiv -1 modp$$ contradicting the minimality of $x$ and if $l=0$ then we get $p=2$ which is not possible as $m$ is odd. Now if $k$ is odd then we get $$2^l \equiv 1 modp$$ contradicting the minimality of $y$ so we get $l=0$ but that will imply $2^m=kx$ but $k$ is odd So a contradiction. Now if $2^m=x$ then we have $2^m>m>p>p-1$ so let $2^m=s(p-1)+t$ with $0 \leq t < p-1$ then we get $$-1 \equiv 2^{2^m} =2^{s(p-1)+t} \equiv 2^t modp$$ contradicting the minimality of $2^m$.So any arbitrary prime factor $p$ of $m$ cannot divide $2^{2^m}+1$ so gcd$(m, 2^{2^m}+1)=1$. Similarly we will get gcd$(n, 2^{2^n}+1)=1$, so $m$ divides $2^{2^n}+1$ and $n$ divides $2^{2^m}+1$. Now without the loss of generality let us assume that $m>n$ and let $q$ be any arbitrary prime factor of $n$ then $q$ divides $2^{2^m}+1$ and so $$2^{2^m} \equiv -1 modq$$. Now let $a$ be the smallest positive integer such that $$2^a \equiv -1 modq$$ and $b$ be the smallest positive integer such that $$2^b \equiv 1 modq$$ then again proceeding in the same way as done above we can prove that there are no solutions with $m, n>1$. $(m, n)=(1,1)$ gives a solution, $n=1$ gives $m$ divides $5(2^{2^m}+1)$ again we can show that $m$ cannot divide $2^{2^m}+1$ so we get $m=5$ giving $(m,n)=(5,1)$ as a solution and similarly $(m,n)=(1,5)$ as a solution.
$$$$Is My Solution Correct????
 A: What you've done looks correct to me, except for one very minor point. About two-thirds of the way down, you wrote
$$2^m>m>p>p-1 \tag{1}\label{eq1A}$$
The $m \gt p$ part should be $m \ge p$ as you could possibly have $m = p$.
FYI, a Fermat number is of the form
$$F_n = 2^{2^{n}} + 1$$
where $n$ is a non-negative integer. Thus, you're asking about the cases where $mn \mid F_{m}F_{n}$. Also, in the Factorization of Fermat numbers section, it says

Édouard Lucas, improving the above-mentioned result by Euler, proved in $1878$ that every factor of Fermat number $F_{n}$, with $n$ at least $2$, is of the form $k\times 2^{n+2}+1$ (see Proth number), where $k$ is a positive integer.

The result above immediately shows that any prime $p \mid n$, you have $p \not\mid F_{n}$ since it would need to be larger than or equal to $2^{n+2}$ but $2^{n+2} \gt n \ge p$. Similarly, for your last case of $m \gt n$ and prime $q \mid n$, you have that the smallest possible factor $\gt 1$ of $F_{m}$ being $2^{m+2}$ means $q \not\mid F_{m}$.
