Find all positive integers $n$ for which $1372n^4 - 3 $ is an odd perfect square. Find all positive integers $n$ for which $1372\,n^4 - 3$ is an odd perfect square.


I tried $\bmod ,4,5,7$ and failed. Next, I used Vieta’s Theorem and failed again.
Any hints, please. Thank you very much!
Edit number and parity already. Sorry for typo
Edit 2 : This question is related to this question.
The equation $y^2=1372x^4-3$ has only one positive integral solution for $x$ and $y$ at which is found at $(1,37)$.
We can use the general technique in this answer https://mathoverflow.net/a/338108 to convert your quartic into Weierstrass form and then we can use MAGMA to find all integral points on the curve.
Step 1: Quartic to Cubic (Weierstrass form)
$y^2=1372x^4-3$ can be transformed into $Y^2=X^3-4116X$ using $X:=1372x^2$ and $Y:=1372xy$ via the steps below
Take
$$y^2=1372x^4-3$$
Multiply both sides by $1372^2x^2$
$$1372^2x^2y^2=1372^3x^6-3\times1372^2x^2$$
$$(1372xy)^2=(1372x^2)^3-(3\times1372)(1372x^2)$$
$$Y^2=X^3-4116X$$
Step 2: Search for Integral Points
Then using MAGMA (An online version is here for you to confirm my work for yourself: http://magma.maths.usyd.edu.au/calc/) we can run the following two lines of code to find all of the integral points on our curve:
E := EllipticCurve([0,0,0,4116,0]);
IntegralPoints(E);

And we get the result: $(0 : 0 : 1)$ which tells us that the only one solution exists (the one that we found manually $(1,37)$).
Alternatively: Easier Solution
We could also run the following to get this answer directly (I realized this command existed after doing the work above, but it confirms the same answer).
IntegralQuarticPoints([1372, 0, 0, 0, -3]);

which gives the only positive output as $[ 1, 37 ]$
Let:
$1372n^4-3=(2k+1)^2$
$343\times4\times n^4=4(k^2+k+1)$
Or:
$343 n^4=k^2+k+1$
Now for $k=18$ we have:
$18^2+18+1=343$
This is the only posibility which gives $n=1$.