Can a composite number $3\cdot 2^n + 1$ divide a Fermat number $2^{2^m}+1$?

In OEIS entry A204620, there is a question (by Arkadiusz Wesolowski) about whether composite numbers of the form: $$3\cdot 2^n + 1$$ can be divisors of a Fermat number, i.e. of a number of the form $2^{2^m}+1$. The question is "answered" by a comment by myself where I quote Morehead's 1906 paper (see Links section on OEIS entry). In it, he states:

It is easy to show that composite numbers of the forms $2^\kappa \cdot 3 + 1,\ 2^\kappa \cdot 5 + 1$ can not be factors of Fermat's numbers.

However, to be sure, is that really easy to see? Or can it be proved in a "hard" way?

Sorry, I wasn't able to complete the proof. I'm sure there are much better ideas, probably involving quadratic residues. I'll post what I have so far just because I thought about it for a few hours, and maybe it'll help someone else answer it.

One can use the fact that every factor of a Fermat number $F_m$ for $m>2$ is of the form $k\cdot 2^{j}+1$, where $k$ is odd and $j>1$. For non-unit factors $j$ must be greater than $0$. (See https://en.wikipedia.org/wiki/Fermat_number#Factorization_of_Fermat_numbers .)

If $3\cdot 2^n+1=ab$ divides $F_m,$ then both $a,b$ must be of the form above. So let $k_1, k_2$ be odd natural numbers and let $j_1,j_2>0$ such that:

$$3\cdot2^n+1=(k_1 \cdot 2^{j_1}+1)(k_2\cdot 2^{j_2}+1)=k_1k_22^{j_1+j_2}+k_12^{j_1}+k_22^{j_2}+1.$$

Assume $j_1\le j_2$. Subtract $1$ and divide by by $2^{j_1}$ on both sides (which is strictly less than $2^n$) to get

$$3\cdot 2^{n-j_1} = k_1k_2 2^{j_2}+k_1+k_22^{j_2-j_1}.$$ This leads to a contradiction with $k_1$ being odd unless $j_2=j_1=j.$ So $$3\cdot 2^{n-j}=k_1k_2 2^j +k_1+k_2.$$

This leads to $$k_1k_2 < 3\cdot 2^{n-2j}$$ and therefore we conclude that $k_1=k_2=1$ or $n-2j\ge 1.$ The case $k_1=k_2=1$ leads to contradiction because you will end up with $3\cdot 2^{n}=2^{n+1}+2^n=2^{2j}+2^{j+1},$ so $j=1=n,$ but $6+1=7$ is not a square. On the other hand, $n-2j\ge 1$ implies $(k_1+k_2)$ is an odd integer times $2^{-j}:$ $$3\cdot 2^{n-2j}=k_1k_2+(k_1+k_2)2^{-j}.$$

There is a short but not completely elementary proof for even $n$ in Robinson's "A Report on Primes of the Form $k\cdot2^n+1$ and On Factors of Fermat Numbers."