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!

Generalization. Let $a,b$ be integers. If there exists consecutive integers $c,d$ such that $a-b=a^2c-b^2d$, then $|a-b|$ is a perfect square.
Proof. If $c=d+1$, we have $a-b=a^2(d+1)-b^2d=(a-b)(a+b)d+a^2$, so $$a^2=(a-b)(1-d(a+b))$$ Now let $g=\text{gcd}(a-b,1-d(a+b))$. We have $g^2|a^2$, so $g|a$. Now we have $g|b$, and we have $g|1$, so $g=1$.
Since $a-b$ and $1-d(a+b)$ are coprime and their multiple is a perfect square, we are done.
The case $c+1=d$ is handled similarly. $\blacksquare$.
