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 $[ 1, 37 ]$
In 1923, Hardy and Littlewood [Some problems of “partitio numerorum”, III,
Acta Math., 44 (1923), pp. 1-70] conjectured that all sufficiently large integers $n$ can be written in the form $n=p+a^2+b^2$
where $p$ is a prime, and $a,b$ are integers. In 1959–1960, Linnik [Hardy–Littlewood problem on the representation as the sum of a prime and two squares,
Dokl. Akad. Nauk SSSR, 124 (1959), pp. 29-30 and
An asymptotic formula in an additive problem of Hardy and Littlewood,
Dokl. Akad. Nauk SSSR, 24 (1960), pp. 629-706] confirmed this conjecture.
See also Hooley, On the representation of a number as the sum of two squares and a prime.
Best Answer
Any pair of twin primes will work, because $(n+1)(n-1)=n^2-1$. Those are the only possible solutions because the problem is equivalent to saying $n^2-1=(n+1)(n-1)$ is a product of two primes, so those two primes must be $n-1$ and $n+1$.