Trying to prove pells equation doesn't work with square numbers

Hello, I am trying to prove that for pells equation "$x^2 - ny^2 = 1$" when n is a perfect square the equation has no solutions. Given that all x, y and n are Natural numbers not including zero.
So far I have got this:
$x^2 - m^2y^2 = 1$
$x^2 - (my)^2 = 1$
$(x + my) (x - my) = 1$ (difference of two squares)

Hello, I am trying to prove that for pells equation "$x^2 - ny^2 = 1$" when n is a perfect square the equation has no solutions. Given that all x, y and n are Natural numbers not including zero.
So far I have got this:
$x^2 - m^2y^2 = 1$
$x^2 - (my)^2 = 1$
$(x + my) (x - my) = 1$ (difference of two squares)
$(x + my) = \frac{1}{(x - my)}$

So from here am I right in saying that the left hand side is going to be 2 or higher and the right surely must be 1 or less? I feel like I'm missing a final step or possibly need to word my statement better, any ideas?

You are home and dry on the second to last line: you have the product of two integers equal to one. What are the possible solutions?

So (x + my) and (x - my) would have to both equal plus or minus 1? That not being possible if all x,y and m are integers, is that proof enough in itself?

You now have two sets of equations (one each for +1 , -1) in two unknowns. Solve them. x+my=1 and x-my=1 implies x=? and y=?

I suppose it would imply x=1 and y=0 but that's not possible because y can't be 0