In a table of fundamental solutions $f_1(x,y)$ to Pell equations,
$$x^2-dy^2=1\tag1$$
with $d<110$, two will stand out,
$$(U_{61})^6 = \big(\tfrac{39+5\sqrt{61}}{2}\big)^6 = x+y\sqrt{61} =1766319049+226153980\sqrt{61}$$
$$(U_{109})^6 = \big(\tfrac{261+25\sqrt{109}}{2}\big)^6 = x+y\sqrt{109} = 158070671986249+15140424455100\sqrt{109}$$
where $U_d$ is a fundamental unit. It is a theorem that if fundamental solution {$u,v$} to,
$$u^2-dv^2=-4\tag2$$
is odd, then $f_1(x,y)$ for $(1)$ is given by $\big(\tfrac{u+v\sqrt{d}}{2}\big)^6 = x+y\sqrt{d}\,$ and, being sixth powers, are generally big. A necessary (but not sufficient) condition is that $d=8m+5$.
In "Large fundamental solutions of Pell's equation", the authors point out that,
$$d=60n'+1$$
$$d=60n'+49$$
have record sizes. Together with the condition $d=8m+5$, then $n'$ must be odd, so,
$$d = 120n+\color{brown}{61}$$
$$d = 120n+\color{brown}{109}$$
hence those two stand-outs are just the first in a rather orderly sequence.
Q: What is the necessary condition such that prime $d = 120n+61$ (or $120n+109$) has odd fundamental solution $u,v$ to $u^2-dv^2=-4$?
I checked that, for primes, the list of $d=120n+61$ with odd solutions {$u,v$} is,
$$d_1 = 61, 181, 421, 541, 661, 1021, 1381, 1621, 1741, 2341, 3061,\dots$$
while those with even {$u,v$},
$$d_2 = 1861, 2221, 3181, 3301, 4021,\dots$$
and it seems that, within a bound, majority belong to the former. So what else (mod constraints?) distinguishes $d_1$ from $d_2$?
Addendum:
For those interested, sequence A107997 is square-free $d$ such that $u,v$ are both odd. Excepting $p = 5$, the rest modulo $120$ fall into eight classes. The best $d<10000$ per class are given below:
$$\begin{array}{|c|c|c|c|c|} \text{#}&\text{Form}&d&\text{digits of}\; u&\tfrac{1}{6}\sqrt{d}\,\ln(\ln(d))\\ 1&120n+13&9013&23&-\\ 2&120n+29&9629&18&-\\ 3&120n+37&8317&23&-\\ 4&120n+53&8933&13&-\\ 5&\color{brown}{120n+61}&8941&\color{brown}{34}&\color{brown}{34.8}\\ 6&120n+77&11117&11&-\\ 7&120n+101&9461&20&-\\ 8&\color{brown}{120n+109}&9949&\color{brown}{36}&\color{brown}{36.9}\\ \end{array}$$
The winners, hands down, are $d=120n+61,\;120n+109$ and they alone approach the website authors' asymptotics. (The sixth class was so poor I extended it to $d<12000$.)