Solutions to the Pell equation $(2x+y)^2-5y^2=4$
Expanding out $(2x+y)^2-5y^2=4$ yields $$x^2+xy-y^2=1\tag{1}$$
It is indeed true that the Fibonacci numbers satisfy eq. $(1)$, for example the pair $(1,1)=(F_1,F_2)$. The next solution should then be $(F_3,F_4)=(2,3)$ which works, since $4+10-9=1$.
If $(x,y)=(F_{2n-1}, F_{2n})$ works, then $(F_{2n+1},F_{2n+2})=(x+y,x+2y)$ should work aswell. This can be proven easily as
$$(x+y)^2+(x+y)(x+2y)-(x+2y)^2=x^2+xy-y^2=1$$