Let $x,y>0$ be integers. Can we find an integer $b>2$ such that at least two solutions exist of the equation $$ x^2 = 1+8\cdot b^y. $$
Initial looking seems to appear that the answer is no. For different values of $b$ I've found there are either there are no solutions (ie $b=9$) or exactly one solution (ie $b=10$), but proving the general case I have stalled in being able to find a way to tackle.
Not looking for full solution, a hint would suffice
On
As requested in comments (this is not a full answer).
Write the equation as $$ (x-1)(x+1)=8\cdot b^y. $$ Common factor of both terms on the left divide $x+1-(x-1)=2$ and it's clear they both must be even, hence $\gcd(x-1,x+1)=2$. So $b^y$ must be split into two coprime factors where one divides $x-1$ and other divides $x+1$, this together with two options for $4\mid x-1,2\mid x+1$ or vice versa gives either $x-1=2c^y,x+1=4d^y$, or $x-1=4d^y,x+1=2c^y$, where $c,d$ are coprime and $b=cd$. Subtracting the two equations in each case and dividing by $2$ gives that either $2d^y-c^y=1$ or $c^y-2d^y=1$. Or just compactly $|c^y-2d^y|=1$.
Now to address the last equation for $y \geq 3$ we can use the result in this MO post by KConrad (mainly Theorem 1.1 of https://www.math.ubc.ca/~bennett/B-Crelle2.pdf). It implies that for each such $y$ the only solution is given by $c=d=1$, hence $b=1$. This does not meet your condition $b>2$ and so we are left with only $y=1$ and $y=2$ cases.
I suspect $b=6$ is the only one.
With your $y = 1,$ we get every odd $x$ with $b = \frac{x^2 - 1}{8}$
With your $y = 2,$ we take all $b$ in the sequence $b_{j+2} = 6 b_{j+1} - b_j, $ that is $ 6, 35, 204, 1189,...$ as we then $x^2 - 8 b^2 = 1.$ Note that the $x_j$ are $17, 99, 577, 336, ...$ and satisfy $x_{j+2} = 6 x_{j+1} - x_j. $
We can fill in the beginnings with $b = 0, 1, 6, 35, 204,...$ matching $x = 1, 3, 17, 99, 577,...$
higher exponents: we can quickly deal with $y=3,$ as then we have $x^2 = 1 + w^3$ and $w=0,1$ are the only integer points.