This question is related to: Are the set of divisors of $N$ closed under multiplication of a linear combination of Fibonacci numbers? (version: 2023-07-11).
Let $(x,y)$ denote the gcd of $x,y$ and $F_n$ denote the $n$-th Fibonacci number (also extended to negative numbers).
Conjecture 1: If $N=aF_m+bF_n=aF_p+bF_q$ and $\pm(mq+np) \notin \{0,1,N\}$ then $(mq+np,N)$ gives a non-trivial divisor of $N$.
Interestingly, for $N=12$ we have
$$ 12=F_4-3F_{-4}=F_8-3F_4 \\ 6=F_4-3F_{-2}=F_8-3F_5 \\ 4=F_7-3F_4=F_1-3F_{-2} \\ 3=F_4-3F_0=F_0-3F_{-2} \\ 2=F_5-3F_2=F_5-3F_1 \\ 1=F_1-3F_0=F_2-3F_0 $$
and for $N = 20$ we have
$$ 20=2F_{-7}-3F_3=2F_{7}-3F_{-3} \\ 10=2F_5-3F_0=2F_6-3F_3 \\ 5=2F_9-3F_8=2F_2-3F_{-2} \\ 4=2F_3-3F_0=2F_5-3F_3 \\ 2=2F_7-3F_6=2F_1-3F_0 \\ 1=2F_3-3F_2=2F_6-3F_5 \\ $$
and for $N = 22$ we have
$$ 22=3F_6-2F_2=3F_6-2F_1 \\ 11=3F_4-2F_{-2}=3F_5-2F_3 \\ 2=3F_0-2F_{-2}=3F_3-2F_3 \\ 1=3F_1-2F_1=3F_2-2F_1. $$
For $N=12$, we have $mq+np=4 \cdot 4+8 \cdot (-4)=-16.$ We have $(12,-16)=4$.
For $N=6$, we have $mq+np=4 \cdot 5+(-2) \cdot 8=4.$ We have $(4,6)=2$.
For $N=4$, we have $mq+np=7 \cdot (-2)+1 \cdot 4=-10.$ We have $(4,-10)=2$.
For $N=20$, we have $mq+np=-7 \cdot (-3)+7 \cdot (3)=42.$ We have $(20,42)=2$.
For $N=22$, we have $mq+np=6 \cdot 1+6 \cdot 2=18.$ We have $(22,18)=2$.
So, the conjecture yields non-trivial divisors of $N$ for these examples.
Is the conjecture true $\forall N \in \mathbb{Z}^{+}$?
Update (Jul 13, 2023): The counterexample given in the answer by @Sil shows that this conjecture 1 is false.
Conjecture 2: If $\exists a,b,m,n,u,v,p,q \in \mathbb{Z}$ such that $N=aF_m+bF_n=aF_p+bF_q, d|N = aF_u+bF_v, \pm(mq+np) \notin \{0,1,N\}$ then $(mq+np,N)$ gives a non-trivial divisor of $N$.
Is the conjecture true $\forall N \in \mathbb{Z}^{+}$?
No.
For example $$ 2=1\cdot F_5-3\cdot F_1=1\cdot F_3-3\cdot F_0 $$ and so with $(a,b)=(1,-3)$, $(m,n,p,q)=(5,1,3,0)$ and $N=2$ you have $$ mq+np=5\cdot 0+1\cdot 3=3 \not\in \{0,1,2\} $$ and yet $$ (mq+np,N)=(3,2)=1 $$ is a trivial divisor of $N$.