Generalizing the Pell equation $x^2-9\times89y^2 = 1$? 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.
 A: 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.
A: 
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 $3$?

Since $d=12n+5\equiv 2\pmod 3$, $$p^2-dq^2=-1\implies p^2+q^2\equiv 2\pmod 3\implies p^2\equiv q^2\equiv 1\pmod 3$$
It follows from this that
$$p^2+dq^2\equiv 1+2\cdot 1\equiv 0\pmod 3$$
