Let's determine the pairs of positive integers $(n, k)$ for which $\left(2^{2^{n}}+1\right)\left(2^{2^{k}}+1\right)$ is divisible by $n k$.
Solution. Let $p$ be any prime greater than 2. Consider the numbers

$$
2^{2^{1}}, 2^{2^{2}}, 2^{2^{3}}, \ldots
$$

These numbers are successive squares of each other, since

$$
\left(2^{2^{n}}\right)^{2}=2^{2^{n+1}}
$$

If any of these numbers has a remainder of -1 when divided by $p$, then the next one will have a remainder of $(-1)^{2}=1$, and all subsequent ones will also have a remainder of 1. This implies that the remainder -1 can occur at most once. Suppose it does occur, i.e., $2^{2^{m}} \equiv-1(\bmod p)$, or in other words, $p \mid 2^{2^{m}}+1$. We will show that in this case $m=1$.

Assume $m>1$. Let $p$ be a prime factor of $n$. Clearly, $p \neq 2$, since $\left(2^{2^{n}}+1\right)\left(2^{2^{k}}+1\right)$ is odd. Since $p \leq n \leq k$, neither $2^{2^{n}}+1$ nor $2^{2^{k}}+1$ can be divisible by $p$, but then their product cannot be divisible by $p$ either, which is a contradiction; hence, $n=1$ must hold. Suppose $k>1$, and one of its prime factors is $p$. Similarly to the previous case, $p>2$, and since $p \leq k$, $p$ does not divide the factor $2^{2^{k}}+1$. Therefore, $p|k|\left(2^{2^{n}}+1\right)\left(2^{2^{k}}+1\right)$ is only possible if $p \mid 2^{2^{n}}+1=5$, so - since every prime factor of $k$ is relatively prime to $2^{2^{k}}+1$ - $k \mid 5$.

Thus, the possible pairs (considering the order) are: $(1,1),(1,5),(5,1)$, all of which satisfy the conditions of the problem.