These are the only solutions if we impose the condition that $y,x>0$, for example. Otherwise, we can clearly find others (like $(-y,-x)$ or $(-x,y)$ for instance). Notice that if $(y,x)$ is a solution, so is $(x,y-ax)$. Since $y,x>0$, the sum of the components decreases when we perform this procedure, and all the solutions are obtained from minimal solutions (i.e. positive solutions which cannot be “decreased” through this procedure without making a component $\le 0$) by applying the inverse transformation. It thus suffices to prove that the only minimal solution is $(y,x)=(a,1)$. This, however follows immediately: since $(y,x)$ is minimal, it must be that $y-ax\le 0$. If we were to have $y-ax<0$, then we would also have $|y^2-ayx-x^2|=|y(y-ax)-x^2|>1$, contradiction. This shows that $y-ax=0\Rightarrow x=1\Rightarrow y=a$, and we’re done.