Those continued fractions are periodic eventually. Therefore, there exists such pair.

Let $x_1$ be an irrational root of a integer coefficient. Then the continued fraction of $x_1$ is eventually periodic. Now let us consider how $f$ act on continued fractions. Let $x=[p;q,r,s,\ldots]$ be an irrational continued fraction. If $p \ge 2$, the $f(x) = [0; p-1,q,r,s,\ldots]$. If $p =1$, then $f(x) = [q;r, s, \ldots]$. If $p=0$, and $q \ge 2$, then $f(x) = [0; q-1,r,s ,\ldots]$ If $p=0$ and $q=0$, then $f(x) = [r;s, \ldots]$. Now, we will study how $f$ can be finitely applied so that the continued fractions are shifted to the left. When $p \ge 2$, then through $[0; p-1,q,r,s, \ldots]$, and then applying $f$ $p-2$ times we have $[0;1,q,r,s, \ldots]$ and then one more $f$ obtains $[q;r,s,\ldots]$. When $p=1$, then one application of $f$ yields $[q;r,s,\ldots]$ and then $q$ times applyiing $f$ to get $[r;s, \ldots]$. In any case, we can apply $f$ finitely many times to have continued fraction shifted by $1$ or $2$ steps to the left. Therefore, we will have at some $j$, $x_j$ with totally periodic. And then with finitely many application of $f$ again, the same continued fraction $x_\ell$.