I. The fundamental solution to the Pell equation, $$x^2-61y^2=1$$
will stand out being "largish" as it is the $\color{blue}{6\text{th}}$ power of a fundamental unit $U_d$,
$$(U_{61})^6 = \big(\tfrac{39+5\sqrt{61}}{2}\big)^6 = x+y\sqrt{61} =1766319049+226153980\sqrt{61}$$
A necessary (but not sufficient) condition is $d=8n+5$.
II. Likewise, the fundamental solution to,
$$x^2-9\times89y^2=1$$
turns out to involve a $\color{blue}{4\text{th}}$ power,
$$(U_{89})^4 = (500+53\sqrt{89})^4 = x+3y\sqrt{89} = 500002000001+3\times 17666702000\sqrt{89}$$
More generally,
Q. Is it true that given fundamental solution $p,q$ of the negative Pell equation, $$p^2-dq^2 = -1$$ if we define $x,y$ as, $$(p+q\sqrt{d})^4 = x+3y\sqrt{d}$$ then $x,y$ is the fundamental solution of, $$x^2-9dy^2 = 1$$ for all prime $d = 12n+5$? Equivalently, since $$(p^2 - d q^2)^4 = \big(p^4 + 6 d p^2 q^2 + d^2 q^4\big)^2 - d \big(4 p q (p^2 + d q^2) \big)^2=1$$ then for such $d$, is $p^2 + d q^2$ a multiple of $3$?
P.S. There is similar behavior for
$$x^2-49dy^2 = 1$$
but now involves a $\color{blue}{24\text{th}}$ power. For example, the fundamental solution to,
$$x^2-49\times13y^2 = 1$$
is,
$$(U_{13})^{24} = \big(\tfrac{3+\sqrt{13}}{2}\big)^{24} = x+7y\sqrt{13} =1419278889601 +7\times56233877040\sqrt{13}$$
though its sequence of primes $d = 5, 13, 61, 157, 181, 397,\dots$ is harder to characterize compared to section II.
Consider the case where $d \equiv 5 \bmod 12$ is prime. From $p^2 - dq^2 = -1$ we obtain that $p, q \equiv \pm 1 \bmod 3$. Since $(p+q\sqrt{d})^2 = p^2 + dq^2 + 2pq\sqrt{d}$, the square of the fundamental unit is not congruent to an integer modulo $3$. Its 4th power, however, is, as you can check by direct calculation or by observing that since $3$ is inert in the quadratic number field ${\mathbb Q}(\sqrt{d})$, its residue class field has order $4$. This result does not change if you allow half-integers, i.e., if you consider the fundamental solution of $p^2 - dq^2 = -4$, since the cube of such a unit has integral coordinates.