EDIT: Dror answered this to my complete satisfaction. The simple continued fraction for $$ \sqrt{n^2 + 1}$$ has length exactly exactly the same, no dependence on $n$!!!!! No lower bound of the type I wanted is possible.
The adjacent reduced indefinite binary quadratic forms in the complete cycle are
$$ \langle 1, \; 2 n, \; -1 \rangle $$
$$ \langle -1, \; 2 n, \; 1 \rangle $$
$$ \langle 1, \; 2 n, \; -1 \rangle $$
for a cycle length of 2.
EDIT: evidently Schinzel studied these "sleepers, creepers, jeepers, etc." with known very short continued fraction lengths. So Dror had explicitly asked me whether I wanted to disregard those but I could not recognize the terminology and gave a misleading answer. So my method may be useless, but the revised question stands. And the person who knows how to correctly describe the question is Dror Speiser.
ORIGINAL QUESTION: This question began with Dror Speiser's answer to