<s>[INST] 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. 
 [/INST] If you want the ratio $(a^2+b^2-1)/(ab)$ to be equal to $a$, choose $b=a^2-1$.
Then the numerator is $$a^2+b^2-1=a^2+(a^2-1)^2-1=a^4-a^2=a^2(a^2-1),$$
which divided by the denominator $ab=a(a^2-1)$ gives the result $a$.
This answers in the affirmative the question of the "update" of the OP, since here $a$ can be any $m \ge 2$.
I noticed after posting this that Will Jagy mentions the choice $a=r,b=r^2-1$ in a comment. I'll delete this if asked to...
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