Newton's method for square roots 'jumps' through the continued fraction convergents
First some experimentation might be in order. I wrote up a program that could check for chains like this:
program nr
implicit none
integer, parameter :: ik16 = selected_int_kind(38)
integer, parameter :: N = 200
integer(ik16) p(N), q(N)
integer D
integer sqD
integer r, s, a
integer i, j, k
integer(ik16) e, b, c
write(*,'(a)',advance='no') 'Enter the value of D:> '
read(*,*) D
sqD = sqrt(D+0.5d0)
r = 0
s = 1
a = (sqD+r)/s
p(1) = a
q(1) = 1
r = a*s-r
s = (D-r**2)/s
a = (sqD+r)/s
p(2) = a*p(1)+1
q(2) = a
do i = 3, N
r = a*s-r
s = (D-r**2)/s
a = (sqD+r)/s
p(i) = a*p(i-1)+p(i-2)
q(i) = a*q(i-1)+q(i-2)
if(p(i) <= p(i-1) .OR. q(i) <= q(i-1)) exit
end do
write(*,'(*(g0))') 'There are ',i-1,' convergents computed'
outer: do j = 1, 5!(i-1)/2
write(*,'(*(g0))') 'Starting convergent: ', j
e = p(j)
b = q(j)
k = j
do
c = e**2+D*b**2
if(c <= e) exit
b = 2*e*b
e = c
do k = k+1, i-1
if(b == q(k)) then
if(e == p(k)) then
write(*,'(*(g0))') 'Matching convergent: ', k
exit
else
write(*,'(a)') 'Matched denominator but not numerator'
stop
end if
else if(b < q(k)) then
write(*,'(*(g0))') 'Missed convergent: ', k-1
cycle outer
end if
end do
end do
end do outer
end program nr
So it always seemed to hit chains for $D=2$, $D=5$, $D=10$, $D=26$, $D=37$, and $D=50$ but for other values of $D$ sometimes it hit and sometimes it missed. My program has indices $1$ bigger than in the original question, so I'm seeing $n\rightarrow2n$ rather than $n\rightarrow2n+1$. Even the infamous $D=61$, which has a pretty large nontrivial solution to Pell's equation, has a chain. When another convergent is hit, it always seems to be exactly $2n$, never $2n-2$ or $2n+2$. It can't be $2n-1$ or $2n+1$ because Newton-Raphson approaches roots from above.
To prove a chain once found, it's probably useful to compare to solutoins to Pell's equation. Probably writing out the solutions in exponential form would lead to proof for a given chain.
EDIT: Just after posting it occurred to me that $$\begin{align}(p^2+Dq^2)^2-D(2pq)^2 & =p^4+2Dp^2q^2+D^2q^4-4Dp^2q^2\\ & =p^4-2Dp^2q^2+D^2q^4\\ & =(p^2-Dq^2)^2\end{align}$$ So every solution to Pell's equation $p^2-Dq^2=1$ leads to an infinite chain of solutions via Newton-Raphson and if $|p^2-Dq^2|\ne1$, then this metric grows exponentially so that the roots computed by Newton-Raphson will not be convergents.
EDIT: Relevant link to Newton-Raphson and Pell's equation.