$4y^2+y=3x^2+x$ implies that $x-y$ is a perfect square. I need a help to prove the following statement. (Sorry for my bad english).

If $x,y\in\mathbb{Z}$ are solutions of $4y^2+y=3x^2+x$, then $x-y$ is a perfect square.

I've tried to rewrite the equation as $\cfrac{y^2}{x-y}=3(x+y)+1 $ and conclude that $x-y\mid y^2$, but I do not think that's useful. I found two solutions: $(x,y)=(2,-2)$  and $(x,y)=(30,26)$. If I could find a recursive way to build more solutions, I would probably show that $x-y\in\mathcal{Q}:=\big\{ n^2\mid n\in \mathbb{N} \big\}$.
Thanks to everyone which will help me to solve (prove) it.
 A: You are on the right track: $4y^2+y=3x^2+x$ implies
$$(x-y)(3(x+y)+1)=y^2\tag{1}$$
which means that $(x-y)$ divides $y^2$. 
Now assume that $(x-y)$ is not a square. Then there exists a prime $p$ and an odd positive integer $k$ such that the largest power of $p$  which divides $(x-y)$ is $p^k$. Then $p$ divides $y^2$ and therefore it divides also $y$ and $x+y=2y+(x-y)$. 
Since $k$ is odd and the RHS of (1) is a square, it follows that $p$ divides also $(3(x+y)+1)$. Hence 
$$p\mid\gcd(x+y,3(x+y)+1)=1$$
Contradiction!
A: Let $x=y+z$, then $$y^2-6zy-z(3z+1)=0$$ Discriminant must be a perfect square, that is $$9z^2+z(3z+1)=12z^2+z=z(12z+1)=w^2$$ Now its easy to see that $\gcd(z, 12z+1)=1$. Thus $z=x-y$ is a prfect square.
A: Hint:
Let $x-y=p\iff x=p+y$
$$0=3(p+y)^2+(p+y)-(4y^2+y)=3p^2+p(6y+1)-y^2$$
$192-24+1$
The discriminant $=(6y+1)^2+12y^2=48y^2+12y+1$ needs to be perfect square $=q^2$(say) 
$$(2q)^2-3(8y+1)^2=1$$
Now use Pell's equation
A: I had found a way to prove it. Thanks to everyone wich pause your lifes to help me. Here goes the solution:
Solution. From $4y^2+y=3x^2+x$, its easy to find the following equations:
$$\begin{cases} y^2=(x-y)\big(3(x+y)+1\big)\\ x^2=(x-y)\big(4(x+y)+1\big)\end{cases} $$
Lets call $\epsilon$ and $\delta$ P.E.S. (from portuguese: "primos entre sí"), numbers wich $\gcd(\epsilon,\delta)=1$.
Therefore, if we have $a^2=\epsilon \cdot \delta  $ for integers numbers, if $\epsilon$ and $\delta$ are P.E.S. , we have that
 $\epsilon$ e $\delta$ are 
necessarily both squares. 
Note that, $3(x+y)+1 $ e $4(x+y)+1$ are P.E.S. , because 
$$ 4\Big(3(x+y)+1\Big)-3\Big(4(x+y)+1\Big)=1$$
I.e. , the linear combination is $1$. Therefore, $3(x+y)+1 $ and $4(x+y)+1$ 
are perfect squares, wich means that $x-y$ must be to.
A: The equation
$$
4Y^2 + Y = 3X^2 + X
$$.
Let me proof why $X - Y = c$.
where $c$ is a perfect square, note: the values $X$ and $Y$ here are the points $( X, Y )$ on the graph of the equation.
$4Y^2+Y = 3X^2+X$.
make $Y$ the subject of the formula there.
$Y = \frac{-1 \pm \sqrt{48X^2+16X+1}}{8}$.
Now say $X-Y = c$.
So that.
$
\begin{align}
X - \frac{-1 \pm \sqrt{48X^2+16X+1}}{8} = c\\
8X - \frac{-1 \pm \sqrt{48X^2+16X+1}}{1} = 8c\\
8X + 1 \mp \sqrt{48X^2+16X+1} = 8c\\
8X + 1 - 8c = \pm \sqrt{48X^2+16X+1}\\
(8X+1-8c)^2 = 48X^2+16X+1\\
1-16c+64c^2+16X-128cX+64X^2 = 48X^2+16X+1\\
16X^2-128cX+64c^2-16c = 0
\end{align}
$.
This equation would give us the value of the $X$ coordinate for which $X-Y = c$.
$X^2-8cX+4c^2-c = 0$.
solving for $X$ shows that.
$X = \frac{8c \pm \sqrt{64c^2-4(4c^2-c)}}{2}$.
which can be further depressed to.
$X = 4c \pm \sqrt{12c^2+c}$.
if we had put $c = 4$, we would get coordinate point $X = 2 \text{and} 30$ as solution.
the determinant here is $12c^2+c$, were $c$ was defined to be a perfect square, let's assume we are looking for integers values of $X$.
Then $12c^2+c = c(12c+1)$ must also be a perfect square, then the number $(12c+1)$ must be a perfect square also, but as far I can say, this worked only when $c = 4$ where $(12×4+1) = 49$, and $49$ is also a perfect number.
Let's recall that $X^2-8cX+4c^2-c = 0$.
now we want to see the value of $X$ that can make $c$ a perfect square.
put as $4c^2-(8X+1)c+X^2 = 0$.
$c = \frac{8X+1 \pm \sqrt{48X^2+16X+1}}{8}$.
one of these condition is that $(48X^2+16X+1)$ must be a perfect square.
we are still limited to only two values that can fit this $X = 2 \text{and} 30$
