If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem. 

If x, y are nonzero natural 
  numbers with $x>y$ such that
  $$x-y = 5y^2 - 4x^2,$$
  prove that $x - y$ is perfect square.

What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) + y^2$$
$$x - y + 4(x-y)(x+y) = y^2$$
$$(x-y)(4x+4y+1) = y^2$$
So $4x+4y+1$ is a divisor of $y^2$.
I also take into consideration that $y^2$ modulo $4$ is $0$ or $1$ (I don't know if this can help.)
So how do I prove that $4x+4y+1$ is a perfect square (this would involve $x-y$ - a perfect square)? While taking examples, I couldn't find any perfect square with a divisor that is $M_4 + 1$ and is not perfect square. 
If there are any mistakes or another way, please tell me.
Some help would be apreciated. Thanks!

Solve, 
$$x - y=5y^2 - 4x^2\tag1$$
hence,
$$x = \frac{-1\pm\sqrt{1+16y+80y^2}}{8}$$
Let,
$$1+16y+80y^2 = \big(\tfrac{2p}{q}y-1\big)^2$$
Expand and factor to get,
$$y = \frac{pq+4q^2}{p^2-20q^2}$$
with the relevant $x$ as,
$$x = \frac{pq+5q^2}{p^2-20q^2}$$
and $x,y$ will be integers if $p,q$ satisfy the Pell equation,
$$p^2-20q^2 = 1$$
One can then see that $x>y$ and,
$$x-y = \frac{q^2}{p^2-20q^2}=q^2$$
The first few $p,q$ are,
$$9, 2 \\161, 36 \\ 2889, 646$$
and these yield $x,y$,
$$38, 34 \\ 12276, 10980 \\ 3952874, 3535558$$
and so on, consistent with the numerical search made in the comments.
