Solution:
Put $A = a - 1$ and the equation becomes $A (2A + 1) = b (3b + 1)$. Let $d$ be the greatest common divisor of $A$ and $b$. Put $A = d x$, $b = d y$. Then $x (2 d x + 1) = y (3 d y + 1)$. Since $x$ and $y$ are coprime, $x$ must divide $3 d y + 1$. So put $3 d y + 1 = n x$. Then $2 d x + 1 = n y$. Solving for $x$ and $y$ in terms of $n$ and $d$ we get $x = \frac{n + 3 d}{n^{2} - 6 d^{2}}$, $y = \frac{n + 2 d}{n^{2} - 6 d^{2}}$.

So we would certainly be home if we could show that there were infinitely many solutions to $n^{2} - 6 d^{2} = 1$. It is not hard to find the first few: $1^{2} - 6 \cdot 0^{2} = 1$, $5^{2} - 6 \cdot 2^{2} = 1$, $49^{2} - 6 \cdot 20^{2} = 1$. We notice that $49^{2} = 2 \cdot 5^{2} - 1$, so we wonder whether $n = 2 \cdot 49^{2} - 1$ might be another solution and indeed we find it gives $d = 1960 = 2 \cdot 49 \cdot 20$. This suggests we try $(2 n^{2} - 1)^{2} - 6 (2 n d)^{2} = 4 n^{4} - 4 n^{2} + 1 - 24 n^{2} d^{2} = 4 n^{2} (n^{2} - 6 d^{2} - 1) + 1 = 1$. So there are indeed infinitely many solutions to $n^{2} - 6 d^{2} = 1$ and we are done.