A simple Google search lead me to a chapter in the proceedings Public-Key Cryptography and Computational Number Theory edited by Kazimierz Alster, Jerzy Urbanowicz, Hugh C. Williams, which contains some results and references which might be interesting in connection with your question, in particular it lists some known results, which I have copied below.

While the Fermat numbers $F_m$ have long been of great interest to mathematicians, the generalized Fermat numbers
$$F_m(a,b)=a^{2^m}+b^{2^m}, \qquad \gcd(a,b)=1$$
and the more special case $b = 1$, were not seriously studied until the 1960s;

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The observations above are explained by the results in this subsection; they were recently published by I. Jiménez Calvo [38].

**Theorem 3.1** (Jiménez Calvo). Let $p=k\cdot 2^n+1$ be a prime, where $k$ is odd and $n=n'2^l$, with $n'>3$ odd.
If $p$ divides the Fermat number $F_m=2^{2^m}+1$, then it also divides the generalized Fermat number
$$F_{m-l}(k,1)=k^{2^{m-1}}+1.$$

To put the next result of Jiménez Calvo into perspective, we quote from [9] the following generalization of the Euler-Lucas theorem (Theorem 2.1):

**Theorem 3.2** (Björn and Riesel). Let $p=k\cdot 2^n+1$ be a prime, where $k$ is odd.
Suppose that $p\mid F_m(a,b)$ and $u \equiv a/b \pmod b$ is a $2^t$-th power residue but not a $2^{t+1}$-th power residue $\pmod p$.
Then $n=m+t+1$.

For a proof, see [9]. A partial converse is given by the following

**Theorem 3.3** (Jiménez Calvo). Let $p=k\cdot 2^n+1$ be a prime, where $k$ is odd.
Let $v:=\operatorname{ord}_2(n)$, and $u$ be such that $2$ is a $2^u$-th power residue but not a $2^{u+1}$-th power residue $\pmod p$.
Furthermore, suppose that $n$, $k$ have a common divisor $d>3$, and if $k' := k/d$, then $2$ is a $k$'-th power residue $\pmod p$.

Then $p\mid k^{2^m}+1$ with $m=n-u-v-1$.

As an immediate consequence (in the case $k' = 1$) we get the following result. By the supplementary laws of quadratic reciprocity we always have $u\ge1$.

**Corollary 3.4.** Let $p=k\cdot 2^n+4$ be a prime, with $k$ odd and $k\mid n$. Then
$$p\mid k^{2^m}+1 \qquad\text{for some}\qquad m\le n-2.$$

[9] Björn, A., Riesel, H. Factors of generalized Fermat numbers. Math. Comp. 67 (1998), 441-446. DOI: 10.1090/S0025-5718-98-00891-6

[38] Jiménez Calvo, I., A note on factors of generalized Fermat numbers. Appl. Math. Lett. 13.6 (2000), 1-5. DOI:10.1016/S0893-9659(00)00045-8

Some other references related to this topic are given there.