Edit:

*Edit: The following argument has an error in the second paragraph. I claimed that the set $\{ D^n\sqrt{D} (\mod 1)\}$ has uncountably many accumulation points, but as Sam Nead points out in the comments, this doesn't follow from $\sqrt{D}$ being irrational. On the other hand, it is true that the set $\{ \sqrt{n}(\mod 1)\}$ is uniformly distributed, and therefore that $\sup \pi(\sqrt{n}) =\infty$. This is because $\sqrt{n+1}-\sqrt{n} \to 0$ as $n\to \infty$. See a paper of Elkies and McMullen for related information on the distribution of this set. The issue is here that the set of accumulation points of $\{ x^n \mod 1\}$ for $x$ irrational may be a countable set. This question is actually discussed in Mendes' paper referred to in the question (I should have looked at it more closely before!). As pointed out in that paper, this can occur for Pisot numbers. Also, the statement is proved by Grisel for many quadratic numbers (as mentioned in a note to Mendes paper). * See also another paper of Grisel.

I think I can show that $sup_n(\pi(x^n))=\infty$ when $x=\sqrt{D}$.

Suppose that $x$ is an irrational, non-quadratic number in the interval $(0,1)$, and $n_i$ is a sequence of integers such that $\underset{i\to\infty}{\lim} \sqrt{n_i} (\mod 1) = x$. Then the claim is that $\pi(\sqrt{n_i})$ is unbounded. Basically, since $x$ has non-repeating continued fraction expansion, one can show that the continued fraction expansions of $\sqrt{n_i}$ have unbounded period. One can see this geometrically, by noting that there is an element of $PGL_2(\mathbb{Z})$ fixing $\pm \sqrt{n_i}$, and fixing the hyperbolic geodesic in $\mathbb{H}^2$ joining these two points. If $a^2-n_i c^2=\pm 1$ is the minimal solution to the Pell equation, then the matrix is: $$\begin{bmatrix} a & cn_i \\ c & a\end{bmatrix}$$ Moreover, the pattern of horoballs which the geodesic penetrates in the Farey packing tells you the periodic continued fraction. When we take $\sqrt{n_i} (\mod 1)$, these geodesics shifted by $-\left[ \sqrt{n_i} \right]$ approach the vertical geodesic through $x$, whose Farey pattern gives the continued fraction expansion for $x$. Therefore $\pi(\sqrt{n_i})$ is unbounded.

Now we have to understand the continued fractions of $\sqrt{D^{2n+1}} = D^n \sqrt{D}$, where $D$ is squarefree. The accumulation points of $D^n \sqrt{D} (\mod 1)$ correspond to the truncations of the base $D$ representation of $\sqrt{D}$. Since $\sqrt{D}$ is irrational, the limit points of these numbers in $[0,1]$ will be uncountable, and therefore contains a non-quadratic irrational $x$ which is a limit of $\sqrt{D^{2n_i+1}} (\mod 1)$, as $n_i\to \infty$. So we see that $\pi(\sqrt{D^{2n_i+1}}) \to \infty$.

2 the argument was in error

Edit: The following argument has an error in the second paragraph. I claimed that the set $\{ D^n\sqrt{D} (\mod 1)\}$ has uncountably many accumulation points, but as Sam Nead points out in the comments, this doesn't follow from $\sqrt{D}$ being irrational. On the other hand, it is true that the set $\{ \sqrt{n}(\mod 1)\}$ is uniformly distributed, and therefore that $\sup \pi(\sqrt{n}) =\infty$. This is because $\sqrt{n+1}-\sqrt{n} \to 0$ as $n\to \infty$. See a paper of Elkies and McMullen for related information on the distribution of this set. The issue is here that the set of accumulation points of $\{ x^n \mod 1\}$ for $x$ irrational may be a countable set. This question is actually discussed in Mendes' paper referred to in the question (I should have looked at it more closely before!). As pointed out in that paper, this can occur for Pisot numbers.

I think I can show that $sup_n(\pi(x^n))=\infty$ when $x=\sqrt{D}$.

Suppose that $x$ is an irrational, non-quadratic number in the interval $(0,1)$, and $n_i$ is a sequence of integers such that $\underset{i\to\infty}{\lim} \sqrt{n_i} (\mod 1) = x$. Then the claim is that $\pi(\sqrt{n_i})$ is unbounded. Basically, since $x$ has non-repeating continued fraction expansion, one can show that the continued fraction expansions of $\sqrt{n_i}$ have unbounded period. One can see this geometrically, by noting that there is an element of $PGL_2(\mathbb{Z})$ fixing $\pm \sqrt{n_i}$, and fixing the hyperbolic geodesic in $\mathbb{H}^2$ joining these two points. If $a^2-n_i c^2=\pm 1$ is the minimal solution to the Pell equation, then the matrix is: $$\begin{bmatrix} a & cn_i \\ c & a\end{bmatrix}$$ Moreover, the pattern of horoballs which the geodesic penetrates in the Farey packing tells you the periodic continued fraction. When we take $\sqrt{n_i} (\mod 1)$, these geodesics shifted by $-\left[ \sqrt{n_i} \right]$ approach the vertical geodesic through $x$, whose Farey pattern gives the continued fraction expansion for $x$. Therefore $\pi(\sqrt{n_i})$ is unbounded.

Now we have to understand the continued fractions of $\sqrt{D^{2n+1}} = D^n \sqrt{D}$, where $D$ is squarefree. The accumulation points of $D^n \sqrt{D} (\mod 1)$ correspond to the truncations of the base $D$ representation of $\sqrt{D}$. Since $\sqrt{D}$ is irrational, the limit points of these numbers in $[0,1]$ will be uncountable, and therefore contains a non-quadratic irrational $x$ which is a limit of $\sqrt{D^{2n_i+1}} (\mod 1)$, as $n_i\to \infty$. So we see that $\pi(\sqrt{D^{2n_i+1}}) \to \infty$.

1

I think I can show that $sup_n(\pi(x^n))=\infty$ when $x=\sqrt{D}$.

Suppose that $x$ is an irrational, non-quadratic number in the interval $(0,1)$, and $n_i$ is a sequence of integers such that $\underset{i\to\infty}{\lim} \sqrt{n_i} (\mod 1) = x$. Then the claim is that $\pi(\sqrt{n_i})$ is unbounded. Basically, since $x$ has non-repeating continued fraction expansion, one can show that the continued fraction expansions of $\sqrt{n_i}$ have unbounded period. One can see this geometrically, by noting that there is an element of $PGL_2(\mathbb{Z})$ fixing $\pm \sqrt{n_i}$, and fixing the hyperbolic geodesic in $\mathbb{H}^2$ joining these two points. If $a^2-n_i c^2=\pm 1$ is the minimal solution to the Pell equation, then the matrix is: $$\begin{bmatrix} a & cn_i \\ c & a\end{bmatrix}$$ Moreover, the pattern of horoballs which the geodesic penetrates in the Farey packing tells you the periodic continued fraction. When we take $\sqrt{n_i} (\mod 1)$, these geodesics shifted by $-\left[ \sqrt{n_i} \right]$ approach the vertical geodesic through $x$, whose Farey pattern gives the continued fraction expansion for $x$. Therefore $\pi(\sqrt{n_i})$ is unbounded.

Now we have to understand the continued fractions of $\sqrt{D^{2n+1}} = D^n \sqrt{D}$, where $D$ is squarefree. The accumulation points of $D^n \sqrt{D} (\mod 1)$ correspond to the truncations of the base $D$ representation of $\sqrt{D}$. Since $\sqrt{D}$ is irrational, the limit points of these numbers in $[0,1]$ will be uncountable, and therefore contains a non-quadratic irrational $x$ which is a limit of $\sqrt{D^{2n_i+1}} (\mod 1)$, as $n_i\to \infty$. So we see that $\pi(\sqrt{D^{2n_i+1}}) \to \infty$.