Show that $x^2-dy^2 = -2$ with $d = m^2+2$ has infinitetly many integer solutions

I want to show that for $d= m^2+2$ the equation

$$x^2-dy^2 = -2$$

has infinitetly many integer solutions. By trying out one can see that $x=\pm m, y=\pm 1$ are solutions, but how do we know that there exist infinitely many solutions?

I have already studied the similar equation $x^2 -dy^2=1$ and shown that for that a fundamental solution is given by $(m^2+1, m)$.

If $$d\in\Bbb N$$ is not a square, and $$x^2-dy^2=k\tag1$$ has some integer solution for $$k\ne 0$$ then $$(1)$$ has infinitely many integer solutions. This is because Pell's equation $$x^2-dy^2=1\tag2$$ has infinitely many integer solutions, and if $$(x_1,y_1)$$ is a solution to $$(1)$$ and $$(x_2,y_2)$$ is a solution to $$(2)$$ then $$(x_1x_2+dy_1y_2,x_1y_2+y_1x_2)$$ is also a solution to $$(1)$$.
• Thank you! Do you mean "is also a solution to $(1)$" in the end? – mathcourse Feb 11 '18 at 11:42
• How did you come up with looking at $(x_1x_2 + dy_2y_2, x_1y_2 + y_1x_2)$? – user7802048 Feb 11 '18 at 14:15
• There is a typo; please fix it.. On the last line, it should say $dy_1 y_2$ instead of $d y_2 y_2$. – evaristegd Jun 24 '19 at 23:40