This is a wild guess about odd perfect numbers. Thus you can see it as an exercise and not as a serious conjecture. I add here the MathWorld's reference dedicated to odd perfect numbers.
Question. Is it possible to rule out the existence of an odd perfect number $N$ having the form $$N=\left(2^{2^n}+1\right)\left(2^{2^n-1}+1\right)^4,\tag{1}$$ where $n$ is a positive integer and $2^{2^n}+1$ a prime number? Many thanks.
As was said is just an exercise to know how to discard the existence of such special form of odd perfect numbers. I don't add the motivation of such form (based in speculations when I was exploring different factorizations and hypothesis), but I believe that there aren't odd perfect numbers of such form.
Odd perfect numbers having such special form must have the form $$N = q\bigg(\frac{q+1}{2}\bigg)^4,$$ where $q = 2^{2^n} + 1$ is a Fermat prime.
It follows that the Descartes-Frenicle-Sorli conjecture is true for this odd perfect number, since $\gcd(q, q+1)=1$. Lastly, you know that $$q < \bigg(\frac{q+1}{2}\bigg)^2$$ is true, basically by (independent) results of Brown, Starni and Dris. (This last inequality is, of course, trivially true by inspection.)
No contradictions so far.
Lastly, it is known that $$D(q)D\bigg(\left(\frac{q+1}{2}\right)^4\bigg)=2s(q)s\bigg(\left(\frac{q+1}{2}\right)^4\bigg)$$ where $D(x) := 2x - \sigma(x)$ is the deficiency of $x \in \mathbb{N}$ and $s(x) := \sigma(x) - x$ is the sum of the aliquot divisors of $x$. You might be able to get a contradiction from this last equation, but I have not checked so myself.
Updated (March 25, 2018)
(Please refer to this paper, starting at page 17.) Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$. If Dris conjecture that $q^k < n$ were unconditionally true, and if in addition one could show that the biconditional $$n < q^{k+1} \iff n < \sigma(q^k)$$ holds, then we would have $q^2 < n$.
This would imply that $$q^2 < \bigg(\frac{q+1}{2}\bigg)^2,$$ from which it would follow that $$4q^2 < q^2 + 2q + 1$$ which would result in the contradiction $$0 > 3q^2 - 2q - 1 = q(3q - 2) - 1 \geq 5\cdot(3\cdot{5} - 2) - 1 = 64.$$
An Aside
I believe that Starni have thus proved something stronger than $q^2 < n$. Indeed, he claims to have proven $$n^2 > \frac{1}{2}\cdot{q^{\alpha}}$$ where $$\alpha = \max(\omega(N) - 1, k)$$ and $\omega(N) \geq 10$ is the number of distinct prime factors of $N$.
Thus, I conjecture that indeed you are right and that there are no odd perfect numbers having such special form.