If $\dfrac{4x^2-1}{4x^2-y^2}$ is an integer, then it is $1$ The problem is the following:

If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is
  also an integer, does it implies that $k=1$?

This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equation (if $k$ is a perfect square) and the second is a pell type equation (if $k-1$ is a perfect square). I've tried setting several values of $k$ to get some solutions but i got nothing. I'm starting to think that $k$ must be $1$.
 A: This is a fun problem! Where did you find it?
There are no solutions except for $k=1$. Assume from now on that $k \neq 1$. Since $k$ is clearly odd, it is also not $0$ and we deduce that $k(k-1)>0$. It is convenient to set $M = k(k-1)$. Also, we may assume WLOG that $x$ and $y\geq0$.
As you did, rewrite the equation to
$$ky^2 - 4 (k-1) x^2 = 1$$
or
$$(ky)^2 - k(k-1) (2x)^2 = k.$$
Set $Y=ky$ and $X=2x$ so the equation is 
$$Y^2 - M X^2 = k. \quad (\ast)$$
We will study the equation $(\ast)$. In the end, we will see that there are no solutions with $X$ even and $Y$ divisible by $k$.
The rest of this proof works inside the ring $R:=\mathbb{Z}[\sqrt{M}]$. Note that $M=k(k-1)$ is not square, so this is an integral domain. For an element $\alpha = a+b \sqrt{M} \in R$, set $\bar{\alpha} = a - b \sqrt{M}$. 
Set $\epsilon = (2k-1) + 2 \sqrt{M}$. Note that $\epsilon \bar{\epsilon} = (2k-1)^2 - 4 k(k-1) = 1$, so $\epsilon$ is a unit of $R$. 
Set $\delta = Y+X \sqrt{M}$. Since $\delta$ is a positive real, and $\epsilon>1$, there is some integer $n$ such that $\epsilon^n \leq \delta < \epsilon^{n+1}$. Write $\delta = \gamma \epsilon^n$. Since $\epsilon$ is a unit, $\gamma$ is in the ring $R$ and, by construction, 
$$1 \leq \gamma < \epsilon.$$
We have
$$\epsilon = 2k-1 + 2 \sqrt{k(k-1)} < 2k+ 2k = 4k.$$
So
$$1 \leq \gamma < 4k.$$
But, also $\gamma \bar{\gamma} = \delta \bar{\delta} =k$. So 
$$\frac{1}{4} \leq \bar{\gamma} \leq k.$$
Write $\gamma = U + V \sqrt{M}$. So
$$\begin{matrix}
1 & \leq & U+V\sqrt{M} & < & 4k \\
\frac{1}{4} & \leq & U-V\sqrt{M} & < & k \\
\end{matrix}.$$
Solving for $V$, we have
$$\frac{-k}{2 \sqrt{M}} < V < \frac{2k}{\sqrt{M}}.$$
The LHS is $\approx -1/2$, and the right hand side is slightly larger than $2$. So $V$ is $0$, $1$ or $2$. We break into cases:
$\bullet$ If $V=0$, then the equation $\gamma \bar{\gamma} =k$ gives $U^2 =k$. So $k$ is a square, say $k=m^2$ and $M=m^2(m^2-1)$. We have
$$Y+X \sqrt{M} = m ((2k-1)+2 \sqrt{M})^n = m (2m^2-1 + 2 m \sqrt{m^2-1})^n.$$
An easy induction on $n$ shows that $Y \equiv (-1)^n m \mod m^2$ so, except when $m=1$, we do not have $Y$ divisible by $k=m^2$. Of course, the case $m=1$ corresponds to $k=1$.
$\bullet$ If $V=1$ then $U^2 - k(k-1) = k$ so $U=k$. We have
$$Y+ X \sqrt{M} = (k+\sqrt{M}) \cdot ((2k-1)+2 \sqrt{M})^n.$$
An easy induction on $n$ shows that $X$ is odd.
$\bullet$ If $V=2$ then $U^2 - 4k(k-1) = k$ so $4k^2-3k = U^2$. This gives $64k^2 - 48 k + 9= 64 U^2+9$ or $(8k-3)^2 - (8U)^2 = 9$. The only ways to write $9$ as a difference of squares are $3^2-0^2$ and $5^2-4^2$. The former gives $k=0$, but we saw that $k$ must be odd; the latter gives no solution since $8$ does not divide $4$.
A: Well notice that both the numerator and denominator are squares, so we may factor 
\begin{equation}
\frac{4x^2 - 1}{4x^2 - y^2} = \frac{(2x+1)(2x-1)}{(2x-y)(2x+y)}.
\end{equation}
It is easy to see that $\gcd(2x + 1, 2x - 1) = 1$, so if $k$ is an integer, it must be the case that the denominator each divides one of the linear factors up top. Then there are only two cases for which this is possible: we must have either $y = 1$, or $2x-y = 1$ and $2x+y = (2x + 1)(2x - 1)$, since otherwise, the positive factor below will be larger than either of the numerator factors and thus cannot divide either of them. 
Let's consider the case where $2x - y = 1$ and $2x + y = (2x + 1)(2x -1)$. From the first equation, we have $y = 2x + 1$, so 
$$2x + y = 4x + 1 = 4x^2 - 1,$$
or 
$$4x^2 - 4x - 2 = 0.$$
However, solving that for $x$ shows that $x$ is not an integer, so this case is impossible. Therefore, we are led to the conclusion that $k$ is also an integer when $y = 1$, and in that case, $k = 1$. 
Edit: EuYu pointed out some details that I overlooked. Refer to the comments for the discussion. 
