Show that in general
$$ (2^{2^n}-1) \not \mid (2^{2^n}+1) $$
Note: The author's solution suggests the question is not accurately phrased. That solution suggests it should say "show that $ (2^{2^n}-1) \mid (2^{2^n}+1) $ is not true in general". In this case a single counter-example is sufficient.
We can show $(2^{2^n}-1) \mid (2^{2^n}+1)$ does not hold in general by finding a counter-example.
Consider $n=1$, then $2^n=2$. This gives us $2^{2^n}-1=3$ and $2^{2^n}+1=5$.
Here $3$ does not divide $5$, and so the given statement does not hold in general.
The interpretation of the question, as given, is that $ (2^{2^n}-1) \not \mid (2^{2^n}+1) $ is true for all natural numbers $n$.
By definition, $x + 1\equiv 2 \pmod {x-1}$ for integer any $x \ge 3$.
For natural number $n \ge 1$, we have $2^n \ge 2$, and so $2^{2^n} \ge 4 \ge 3$.
This means, for $n \ge 1$,
$$ 2^{2^n}+1 \equiv 2 \pmod {2^{2^n}-1} $$
The congruence 2 means $2^{2^n}+1 \not \mid {2^{2^n}-1}$.