D

$\because a_{1} \gt a_{2} \gt a_{3} \gt 0$，
$\therefore \frac{1}{{a}_{1}} \lt \frac{1}{{a}_{2}} \lt \frac{1}{{a}_{3}}$，
$\because x_{1}$，$x_{2}$，$x_{3}$都是大于$1$的数，
$\therefore (x_{1}+1)(x_{1}-1) \gt 0$，$a_{2}(x_{2}+1)(x_{2}-1) \gt 0$，$a_{3}(x_{3}+1)(x_{3}-1) \gt 0$，
$\because a_{1}(x_{1}+1)(x_{1}-1)=1$，$a_{2}(x_{2}+1)(x_{2}-1)=2$，$a_{3}(x_{3}+1)(x_{3}-1)=3$，
$\therefore (x_{1}+1)(x_{1}-1)=\frac{1}{{a}_{1}}$，$(x_{2}+1)(x_{2}-1)=\frac{1}{{a}_{2}}$，$(x_{3}+1)(x_{3}-1)=\frac{1}{{a}_{3}}$，
$\because \frac{1}{{a}_{1}} \lt \frac{1}{{a}_{2}} \lt \frac{1}{{a}_{3}}$，
$\therefore (x_{1}+1)(x_{1}-1) \lt (x_{2}+1)(x_{2}-1) \lt (x_{3}+1)(x_{3}-1)$，
$\because (x_{1}+1)(x_{1}-1)=x_{1^{2}}-1$，$(x_{2}+1)(x_{2}-1)={x}_{2}^{2}-1$，$(x_{3}+1)(x_{3}-1)=x_{3^{2}}-1$，
$\therefore x_{1^{2}}-1 \lt {x}_{2}^{2}-1 \lt x_{3^{2}}-1$，
$\therefore x_{1^{2}} \lt {x}_{2}^{2} \lt x_{3^{2}}$，
$\because x_{1}$，$x_{2}$，$x_{3}$都是大于$1$的数，
$\therefore x_{1} \lt x_{2} \lt x_{3}$.
故选：$D$.