I am writing two definitions, the $1$st one is a cover in some sense while the $2$nd one is a lattice:
Definition 1: If $m,n$ are integers bigger than $1$, then the set $$A=\{(x,y) \in \mathbb Q^{\geq 0} \times \mathbb Q^{\geq 0}~:~ mn+1 \mid m^x+n^y \}$$ is said to be cover of $m$ and $n$.
Example: The $5^{th}$ Fermat number is $2^{2^5}+1=641 \times 6700417$. Now $5 \times 2^7+1=641=2^4+5^4$. Therefore \begin{align} &5 \times 2^7+1 \mid 5^4+(2^7)^{4/7} \\ &mn+1 \mid m^x+n^y \end{align} with $m=5,~n=2^7$ and $x=4,~y=4/7$. Thus by definition $(x,y)=(4,4/7)$ covers $(5,2^7)$. Similarly, $(0,2^{5/7})$ belongs to cover of $5$ and $2^7$, because $5 \times 2^7+1$ factors $5^{th}$ Fermat number.
Definition 2: Note $\mathbb Q^2$ is a vector space over $\mathbb Q$. Let $\vec u$ is an "arbitrary" vector in $\mathbb Q^2$ and $\{\vec v, \vec w \}$ is a basis of $\mathbb Q^2$. Then the set $$B=\{\vec u+i \vec v+j \vec w~\mid~ i,j \in \mathbb Z\}$$ is a a lattice in $\mathbb Q^2$.
My question-
I want to show $A \subset B$.
By the above example, $A \ni (4,4/7)=(1,2)+3(1,0)-(0, 10/7)=\vec u+3 \vec v-\vec w \in B$. Note $\{ \vec v, \vec w \}$ forms a basis of $\mathbb Q^2$. Thus, $(4,4/7) \in A \Rightarrow (4,4/7) \in B$. So $A \subset B$ in this case.
Similarly, $A \ni (0,2^{5/7})=(1,-1)+2(-2,2)+(3,11/7) \in B$. Thus $(0,2^{5/7}) \in B$. So $A \subset B$, in this particular case.
General case, take any $(x,y) \in A$. Then since $\{ \vec v, \vec w \}$ is a basis of $\mathbb Q^2$, we can write $(x,y)=a \vec v+b \vec w$ for $a,b \in \mathbb Q$. We don't know whether $a,b$ are integers or not. But we can suitably choose the vector $\vec u \in \mathbb Q^2$ such that $\vec u+a \vec v+b \vec w=(r,s)$ , where $r,s \in \mathbb Z$. Finally, we can "again decompose" $(r,s)$ as $(r,s)=\vec u+i \vec v+j \vec w$ for some suitable $i,j \in \mathbb Z$ and $u,v,w \in \mathbb Q^2$. This implies $(x,y) \in B$.
Thus $A \subset B$. But I didn't require to use the property of $A$.
I seek your opinion.