Let $x^2+y^2+m=kxy$ if we find infinitely many $x,y$ with $(x,y)=1$ satisfy this one we are done. Take $k=m+2$ (we select that because then it is true for $x=y=1$). Then by Vieta Jumping we can see that if $(x,y)$ is a sollution (with $(x,y)=1$) with $y>=x$ then so it is the pair $((m+2)y-x,y)$ and obvious $((m+2)y-x,y)=(x,y)=1$ Some of them is: $(1,1)\rightarrow (m+1,1)\rightarrow ((m+1)(m+2)-1,m+1)=(m^2+3m+1,m+1)\rightarrow ((m+2)(m^2+3m+1)-(m+1),m^2+3m+1)=(m^3+5m^2+6m+1,m^2+3m+1)$

First, if we can find infinitely many positive integers satisfying $x^2+y^2+m=(m+2)xy$, then the problem is solved. Notice however that $(1,1)$ is always a solution. We will prove that if $(x,y)$ is a solution, then so is $(y, (m+2)y-x)$. This can be verified via direct expansion: \begin{align*} y^2+((m+2)y-x)^2 + m =(m+2)y((m+2)y-x) \\ y^2+(m+2)^2y^2 - 2(m+2)xy+x^2+m = (m+2)^2y^2-(m+2)xy \\ x^2+y^2 + m = (m+2)xy \end{align*}This completes the solution.