Numerical evidence suggest that the class number of the quadratic field $x^2=3\cdot2^n+1$ is $O(n)$ while the discriminant is $O(2^n)$.

Here are the class numbers for $n=6 \dots 110$ computed with pari/gp:

1, 2, 1, 2, 1, 4, 1, 3, 2, 4, 1, 1, 1, 2, 3, 1, 2, 1, 2, 2, 2, 4, 48, 1, 3, 2, 1, 4, 2, 2, 1, 8, 4, 4, 4, 5, 1, 2, 6, 35, 3, 80, 25, 2, 4, 24, 4, 4, 12, 8, 2, 1, 24, 12, 2, 8, 8, 1, 8, 4, 13, 104, 4, 2, 1, 8, 4, 8, 400, 2, 4, 1, 1, 4, 2, 2, 2, 4, 4, 2, 10, 80, 2, 16, 2, 16, 2, 372, 4, 32, 4, 46, 8, 6, 8, 12, 6, 1, 4, 4, 4, 4, 8, 4, 12, 4, 8, 8, 4, 72
 

A Monte Carlo factoring algorithm with linear storage possibly might be used to verify for larger $n$, though I don't have working implementation yet.

The motivation is that small class number and the above algorithm might give a divisor of numbers of these form.

I get similar results for $x^2=4\left(3\cdot2^n-1\right)$