For what positive integer values $b,d$ does $(b^2-d)\mid(b^2-1)?$ hold? I am curious about the answer to the following questions: And hope that you can help me

For what positive integer values $b, d$ does
  $$(b^2-d)|(b^2-1)?$$
  hold?

Is it correct that the only solutions are those of the form $(b,1)$? The following leads to an affirmative answer:
The above expression implies $(b+\sqrt d)(b-\sqrt d)\mid(b^2-1)$. This means that either $b+\sqrt d$ or $b-\sqrt d$ divides $b^2-1$. 
Using synthetic division in solving $\frac{b^2-1}{b+\sqrt d}$ and $\frac{b^2-1}{b-\sqrt d}$ I get both the remainer $d-1$. Since we want $d-1=0$ we have $d=1$. So the solution is $(b,d)=(b,1)$.  
Thanks in advance.

This is only a partial answer to your question.
Since we have the identity
$$\frac{b^2 - 1}{b^2 - d} = \frac{b^2 - d}{b^2 - d} + \frac{d - 1}{b^2 - d} = 1 + \frac{d - 1}{b^2 - d}$$
then your question is equivalent to


For what positive integer values $b, d$ does
    $$b^2 - d \mid d - 1$$
    hold?


Now, $b^2 - d \mid d - 1$ implies that $b^2 - d \leq d - 1$, which implies that $b^2 \leq 2d - 1$.
Unfortunately, this does not seem to narrow things that much.
