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

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 $[ 1, 37 ]$