Is $x=2,y=13$ the unique solution? Problem:
Find all positive integers $x$ and $y$ satisfying:
$$12x^4-6x^2+1=y^2.$$
If $x=1, 12x^4-6x^2+1=12-6+1=7,$ which is not a perfect square.
If $x=2, 12x^4-6x^2+1=192-24+1=169=13^2$, which is a perfect square. Thus, $x=2,y=13$ is a solution to the given Diophantine equation.
However, after testing a few more small cases, it seems as if $12x^4-6x^2+1$ can never be a perfect square if $x>2$. I have tried to prove this, but to no avail. Here is the gist of what I considered:
$12x^4-6x^2+1=y^2 \iff 12x^4-6x^2 = (y+1)(y-1)$. Since the L.H.S. is a multiple of $2$, it follows that $2 \mid (y+1)(y-1) \Rightarrow y$ is odd $\Rightarrow y+1$ and $y-1$ are both even.
Hence, $4 \mid (y+1)(y-1) \Rightarrow 4 \mid  12x^4-6x^2 \Rightarrow 4 \mid 6x^2 \Rightarrow x $ is even. Let $x=2m$ and $y+1=2k$, so $12x^4-6x^2=192m^4-24m^2=4k(k-1) \Rightarrow 48m^4-6m^2=k(k-1)$. At this point, I am at a loss of how to continue. We could continue with divisibility arguments, but it seems to be a never-ending process?
Another method I tried was to let $y=x+k$, thus $12x^4-6x^2+1 = x^2 + 2xk+k^2 \iff k^2+2xk-(12x^4-7x^2+1)=0$, which is a quadratic in terms of $k$. However, stuff like the discriminant or sum and product of roots did not seem to yield any important information.
Any hints provided to point me in the right direction will be much appreciated.
 A: This is an addition to my previous answer, which by pointed out by Mike doesn't sufficiently solve the problem.
We use the result, $$3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$$
We have that $\gcd(2a-1,2a+1)=1$ for positive integer $a$ (by Euclidean Algorithm if you need convincing, however this is trivial enough to state).
It follows that $2y-1$ cannot be divisor to $2x-1$ or $2x+1$ without being divisor to the full square term (in other words, if $2y-1 \vert 2x-1,$ then $2y-1 \vert (2x-1)^2$ and similar). Since it also applies the other way around, it follows that one of the following are true:

*

*$2y-1 = 1$ or $2y-1 = 3$

*$2y-1 = (2x-1)^2$ or $2y-1 = 3(2x-1)^2$

*$2y-1 = (2x+1)^2$ (cannot be $3(2x+1)^2$ as we must have $2y+1 > 2y-1$)

For the first, we have $y = 1$ or $y=2.$ No $x$ makes this true, so we can simply ignore this case.
The second case (after making $y$ in terms of $x$ and rearranging)
$2y-1 = (2x-1)^2$ gives us the polynomial $4x^4 + 4x^3 - 7x^2 +2x = 0,$ which has roots $-2, 0, 1/2, 1/2.$ None of these are positive integers.
$2y-1 = 3(2x-1)^2$ gives a polynomial with rational roots $1/2,0$ and the rest are irrational.
$2y -1 = (2x+1)^2$ gives us rational roots (-1/2, 0, 2). 2 is thus an answer!
I may have some cases I overlooked but it seems that this can solve it, albeit in a depressingly bashy way.
A: We try to get a perfect square term on the LHS.
Note that multiplying both sides by 4 keeps the right side as a square:
$$48x^4 - 24x^2 + 4 = (2y)^2$$
Now, completing the square yields: $$3(4x^2 -1)^2 +1 = (2y)^2$$
Try to now find the solutions to $3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$
