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.
 A: 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 $[ 1, 37 ]$
A: The function: $\sqrt{1372n^4 - 3}$ produces one integer only: when $n = 1$, it produces $37$.
For $ n \geq 2$, the function produces decimals.
Therefore, there is no positive integer $n$ such that $1372n^4 - 3$ is a odd perfect square.
This is a brute force function that I tested out on Python - ran the code till $10000$.
A: Just to give another way.
$1372n^4=m^2+3\Rightarrow2n^4\equiv m^2+3\pmod{10}$.
Noting $\mathbb Z/10\mathbb Z=\mathbb Z_{10}$ we have
$$\mathbb Z_{10}^4=\{1,6,5,0\}\Rightarrow2\mathbb Z_{10}^4=\{2,0\}\\\mathbb Z_{10}^2=\{1,4,9,6,5,0\}\Rightarrow\mathbb Z_{10}^2+3=\{4,7,2,9,8,3\}$$
Since $2\mathbb Z_{10}^4\cap(\mathbb Z_{10}^2+3)=\{2\}$ we deduce that modulo $10$ we must have $n=1$ which corresponds to $m=7$ (because of $7^2+3\equiv2\pmod{10}$).
It follows that the only solution is $n=1$.
