Finding conditions such that $4b^2 > a^2 > 3b^2$ and $b \mid (a^2-1)$ imply $b=(a+1)/2$ Consider the set of odd positive integers $a$ and $b$ such that $4b^2 > a^2 > 3b^2$ and $b \mid (a^2-1)$.
Brute-force computation suggests that $a=2b-1$ is the only solution for “most” such $b$, with exceptions:
$$ b \in \{105,153,171,231,253,315,325,333,345,\dots\}.$$
I’m trying to find a finite (and hopefully quite small) set of conditions which eliminates all exceptions.
The first constraint is easy: requiring $3 \nmid b$ reduces the set of exceptions to
$$ b \in \{253,325,425,473,551,595,689,703,713,\dots\}.$$
For $b < 1000$, this is $16$ exceptions, all satisfying
$$1.74 < \frac{a}{b} < 1.90$$
Can anyone suggest other conditions that would eliminate these “rogue” values, so that the complete set of conditions forces $b = (a+1)/2$?

NOTE: I’ve included the “quadratic-residues” and “modular-arithmetic” tags because I imagine those kind of considerations will be involved (or at least would likely be helpful).


EDIT #1: Adding the condition $a \equiv 1\!\!\pmod{4}$ reduces the set of exceptions to
$$b \in \{253,425,595,689,713,737,893,925,979,\dots\}.$$
 A: In order for an odd number $a$ to satisfy $4b^2\gt a^2$, we need $a=2b-(2r+1)$ with $r\ge0$. In order for $a$ to satisfy $a^2\gt3b^2$, we need $r$ to satisfy $2r\lt(2-\sqrt3)b-1$. Since $a^2-1=4b^2-4b(2r+1)+4r(r+1)$, we have $b\mid(a^2-1)$ if and only if $b\mid4r(r+1)$. If $b$ is odd, this reduces to $b\mid r(r+1)$.  So constructing "exceptions" amounts to finding positive values of $r$ for which $r(r+1)$ has an odd divisor $b$ such that $2r\lt(2-\sqrt3)b-1$, which can be rewritten as $2(2+\sqrt3)r\lt b-(2+\sqrt3)$.
If $b$ is an odd divisor of $r(r+1)$ and $r\gt0$, then $b\le r(r+1)/2$. Putting this together with the inequality  $2(2+\sqrt3)r\lt b-(2+\sqrt3)$, we get a quadratic inequality that $r$ must satisfy,
$$r^2-(7+4\sqrt3)r-(4+2\sqrt3)\gt0$$
We find that
$$r\gt{7+4\sqrt3+\sqrt{113+64\sqrt3}\over2}\approx14.445$$
so any search for exceptions must begin with $r\ge15$.
The easiest way to produce exceptions is to let $r=4k+1$ and $b=r(r+1)/2=(4k+1)(2k+1)$, or $r=4k+2$ and $b=r(r+1)/2=(2k+1)(4k+3)$, with $k\ge4$. The restriction $3\not\mid b$ rules out the latter and restricts the former to $r=12k+1$ and $b=(12k+1)(6k+1)$.  I don't see any obvious restriction that would eliminate all exceptions, but maybe I'm missing something.
