This question is a bit involved so thanks for your patience. After a lot of work I could show that each $b$ that satisfies this equation corresponds to another smaller integer say $s$ which satisfies the equation. here are the formulas for the solutions. $$t=s\pm\sqrt{2s^2+1}\tag{1}$$
$$x=3t\pm2\sqrt{2t^2-1}\tag{2}$$
$$y=\pm\sqrt{tx-1}\tag{3}$$
$$b=2x+y\tag{4}$$
$$a=3x+y\tag{5}$$ for nonzero $t$ and $x$. Let $x_+$ and $x_-$ denote the $x$ with the positive sign and negative sign respectively and define $y_+$ and $y_-$ similarly. Use the following rules: If you choose positive $t$ not $1$ you may choose any one of the $x$s, but you must choose $y_+$ if you take $x_+$. If $t=1$ you may take both $x$s and with $x_+$ you must choose only $y_+$. If you take $t$ positive and choose $x_-$ you must choose $y_-$, this last solution repeats the previous solutions but it generates one the others don't when $s=2$ , $t=5$ , $x_-=1$ , $y_-=-2$ , $b=0$ , $a=1$. There are similar rules for negative $t$ but they are just the negatives of the previous solutions. I (think) I could prove that no $s$ can lead to a $b=s$ since all solutions either get smaller or larger but the absolute value always increases except for the case I singled out which leads to $b=0$, which implies that this is the complete set of solutions. I noticed that the values of $b$ increase and at some point they reach a constant ratio which is roughly $\phi=93222358/15994428$ and I used this to derive the function $$f(x)=\phi^{\left(x-11\right)}93222358$$ which is extremely accurate at first and almost exact for $x\ge7$. It gives the $x$-th nonzero value of $b$ where the zeroth value is just $0$. Here's where it gets interesting, I noticed a pattern among the $a$s, that for nonzero even $x$, $a$ can be written in the form $a=c_1^2-1$ and for odd $x$, $a=\ c_2^2+1$. I checked this until $x=15$ and it holds. Also, another reason to believe that $a$ can't be a square is that while I was trying to prove Fermat's last theorem for $n=4$ (where this equation came up), I could reduce the proof to verifying that that no such $a$ can be a square. Of course, I'm a lot more interested now than when I first encountered the equation and I want to prove this pattern.
Best Answer
I think there is a MUCH more elementary way to solve this.
Note that $a$ must be odd. So $a=2n+1$ for some integer $n$.
Solving $(2n+1)^2 = 1+2b^2$ gives $(2n^2+2n) = b^2$ which gives $2(n+1)n$ a perfect square.
This implies that if $n$ is even, then $n+1$ must be an odd square [indeed suppose $p^e$ divides $n+1$ for some odd prime $p$ and some positive integer $e$, then as $p$ does not divide $2(n+1)$ it follows that $e$ must be even] and thus $2n$ an even square $c^2$. So here $a=c^2+1$. And if $n$ is odd then $2(n+1)$ must be an even square $c^2$. So here $a=(2n+2) -1$ which is $c^2-1$.