The only possibilities are $(x_1, x_2) = (c, c)$ for some nonzero integer $c$.

Rewrite the recurrence relation as
$$\frac{2}{x_{n-1}} = \frac{1}{x_n} + \frac{1}{x_{n-2}}.$$
Let $y_n = \frac{1}{x_n}$ for all $n$. This implies
$$y_{n-1} - y_{n-2} = y_n - y_{n-1}.$$
Therefore, $\{y_n\}$ is an arithmetic sequence. Let $y_n = a + n d$ for some constants $a$ and $d$. Then we have
$$x_n = \frac{1}{y_n} = \frac{1}{a + n d}.$$
If $d \neq 0$, then we have $|a + n d| \to \infty$ when $n \to \infty$, and hence $x_n \to 0$. However, $x_n \neq 0$. This shows $x_n$ cannot be an integer for large $n$. So it suffices to consider $d = 0$. In that case $\{x_n\}$ is a constant sequence. In order that there are infinitely many integer terms, the constant must be an integer. So $x_1 = x_2 = c$ for some integer $c \neq 0$. Note that a constant sequence satisfies the recurrence relation since
$$c = \frac{c^2}{2c - c}.$$