Proving conditions on $(r+s)^2 \mid (4r^4+1)$, related to Pell oblongs While working on another problem (Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$), I came across a question which seems to be of [semi-] independent interest.
Conjecture. If $r > s \ge 1$ are integers such that $r$ is odd and $s$ is even and
$$(r+s)^2 \mid (4r^4+1), \qquad(\star)$$
then $(r,s) \in \{(3,2), (21,8), (119,50), (697,288), (2679,910), (4059,1682), \dots\}$.
Note that these are essentially Pell oblongs (i.e., $r$ is http://oeis.org/A001652) and doubles of Pell squares (i.e., $s$ is http://oeis.org/A114619), except for the outlier solution $(2679,910)$, which is the only one maxima gave for $r \le 11000$ (I'm increasing that bound now).
I haven't the foggiest notion how to prove any conditions on $r$ and $s$ satisfying $(\star)$, or to figure out why there are any 'outlier' solutions at all (and, of course, if there are any more). EDIT: For example, it seems that all solutions (including the 'outlier') with $r>3$ satisfy $s/r < 1/2$; can one prove this bound?
Any suggestions, hints, full solutions, etc. greatly appreciated.
 A: [corrected $-$ see edit history for previous attempt]
The conjecture is false: there are infinitely many "Pell" parametrizations,
some with larger values of $s/r$.  For example,
$$
(r,s) = (307470495089672071303, \, 295528756570432706202)
$$
has $s/r \sim .961$.
This was obtained as follows.  Recall that $4r^4 + 1$ factors as
$(2r^2-2r+1) (2r^2+2r+1)$.  Start from the first solution
$(r,s) = (3,2)$, with $2r^2-2r+1 = 13$ and $2r^2+2r+1 = 5^2$.
Instead of generalizing to $2r^2+2r+1 = y^2$, we generalize to
$2r^2-2r+1 = 13y^2$ and $2r^2+2r+1 \equiv 0 \bmod 25$.
This is a Fermat-Pell equation with a congruence condition,
and since we have one solution $(r,y) = (3,2)$ there must be
infinitely many others.  The equation is $x^2-26y^2=1$ with $x=2r-1$,
which must be positive and $1 \bmod 4$ to satisfy the sign and
parity conditions on $n$.  The general solution is
$x + \sqrt{26} \, y = (5 + \sqrt{26})^{4k+1}$ ($k=0,1,2,\ldots$),
and then the ${}\bmod 25$ condition gives $5|k$.  The solution
displayed above comes from $k=5$.
We can obtain further infinite families by iterating 
trick of switching between the $2r^2-2r+1$ and $2r^2+2r+1$ factors,
and by starting from some other solution such as the $r=2679$ "outlier"
or any other solution that a numerical search might find.
