7. Given the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $C_{2}: \frac{x^{2}}{m^{2}}-\frac{y^{2}}{n^{2}}=1(m>0, n>0)$ have the same foci $F_{1}, F_{2}$, where $F_{1}$ is the left focus. Point $P$ is the intersection point of the two curves in the first quadrant, $e_{1}, e_{2}$ are the eccentricities of the curves $C_{1}, C_{2}$, respectively. If $\triangle P F_{1} F_{2}$ is an isosceles triangle with $P F_{1}$ as the base, then the range of $e_{2}-e_{1}$ is $\qquad$.
Notice that $\left|P F_{2}\right|=2 c$, so $\left|P F_{1}\right|=2 a-2 c=2 m+2 c \Rightarrow a-m=2 c \Rightarrow \frac{a}{c}-\frac{m}{c}=2 \Rightarrow \frac{1}{e_{1}}-\frac{1}{e_{2}}=2$ $\Rightarrow e_{2}=\frac{e_{1}}{1-2 e_{1}} \Rightarrow e_{1}\frac{1}{3} \Rightarrow \frac{1}{3}<e_{1}<\frac{1}{2}$, thus $\frac{1}{e_{1}}-\frac{1}{e_{2}}=2 \Rightarrow e_{2}-e_{1}=2 e_{1} e_{2}=\frac{2 e_{1}^{2}}{1-2 e_{1}}=\frac{2}{\frac{1}{e_{1}^{2}}-\frac{2}{e_{1}}}=\frac{2}{\left(\frac{1}{e_{1}}-1\right)^{2}-1}$, since $\left(\frac{1}{e_{1}}-1\right)^{2}-1 \in(0,3)$, so $e_{2}-e_{1} \in\left(\frac{2}{3},+\infty\right)$.