So I came up with $b= a+1$ $\Rightarrow$ $ab=a(a+1) = a^2 + a$
So that:
$a^2+b^2 -1$ = $a^2 + (a+1)^2 -1$ = $2a^2 + 2a$ = $2(a^2 + a)$ $\Rightarrow$
$(a,b) = (a,a+1)$ are solutions.
My motivation is for this follow up question:
(b) With $a$ and $b$ as above, what are the possible values of: $$ \frac{a^2 +b^2 −1}{ab} $$
Update
With Will Jagy's computations, it seems that now I must show that the ratio can be any natural number $m\ge 2$, by the proof technique of vieta jumping.
Update
Via Coffeemath's answer, the proof is rather elementary and does not require such technique.