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 $1$.

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. 
