The required sequences are all constant sequences of positive rational numbers. Clearly, any such satisfies $(*)$.

Let $(a_n)_{n \ge 1}$ be a sequence of positive rational numbers satisfying $(*)$. Then so does $(r a_n)_{n \ge 1}$, where $r$ is any positive rational number. Letting $r$ be the product of the denominators of $a_1$ and $a_2$, we may assume that $a_1$ and $a_2$ are integers. If $a_k$ and $a_{k+1}$ are integers, then so is $a_{k+2}$, as it is a root of the monic polynomial $X^2 - a_{k+1} X + a_k^2 - a_k a_{k+1}$ with integer coefficients. Inductively, the $a_n$ are all integers.

Write $(*)$ in the form
$$(a_{k+1} - a_k)a_k = (a_{k+2} - a_{k+1})a_{k+2}. \quad (**)$$
If $a_1 > a_2$, then $(**)$ forces $(a_n)_{n \ge 1}$ to be a strictly decreasing sequence of positive integers and we reach a contradiction.

If $a_1 < a_2$, then $(a_n)_{n \ge 1}$ is strictly increasing, by $(**)$.
$$a_{k+1} = \frac{a_k^2 + a_{k+2}^2}{a_k + a_{k+2}} > \frac{(a_k + a_{k+2})^2}{2(a_k + a_{k+2})} = \frac{1}{2}(a_k + a_{k+2}) \quad \text{for all } k \ge 1,$$
it follows that $a_{k+1} - a_k > a_{k+2} - a_{k+1}$ for all $k \ge 1$, so the $a_{n+1} - a_n$ form a strictly decreasing sequence of positive integers and we reach again a contradiction.

Finally, if $a_1 = a_2$, then $(**)$ forces $a_n = a_1$ for all $n \ge 1$, so the sequence $(a_n)_{n \ge 1}$ is indeed constant, as desired.