Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. 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. 

$(3,8)$ is a possible solution.
This gives us 24 divides 72, and a value of 3 for (b).
Have you considered that if $ab$ divides $a^2+b^2-1$, then we $ab$ divides $a^2 + b^2 -1 + 2ab$?
This gives us $ab$ divides $(a+b+1)(a+b-1)$.
Subsequently, the question might become easier to work with.
