The answer is yes and moreover, you can identify $F$ and $G$ only using two order statistics: $Z_1$ and $Z_{m+n}$.
As in air's answer, let's consider the case $m=n=1$. Let $H(t)=\Pr[Z_1\le t]$ and $I(t)=\Pr[Z_2\le t]$ be the known cdfs of $Z_1=\min(X,Y)$ and $Z_2=\max(X,Y)$. Again, as in air's answer, we have that $H$ and $I$ are related to $F$ and $G$ by the system of equations:
\begin{align*} H=&1-(1-F)(1-G)=F+G-F\,G\\ I=& F\,G \end{align*}
Thus, since $F=I/G$ by the second equation, we obtain from equation one that $H = \frac{I}{G}+G-I$, or $$ G^2- (H+I) G + I =0$$.
Solving the quadratic equation give us: $G=\frac{(H+I)+\sqrt{(H+I)^2-4I}}{2}$ (the negative root is not consistent with $1\ge H\ge I \ge 0$) and so $F=\frac{2I}{(H+I)+\sqrt{(H+I)^2-4I}}$. That is, we identified $F$ and $G$.
Now consider the general case, $H(t)=\Pr[Z_1\le t]$ and $I(t)=\Pr[Z_{m+n}\le t]$ and so:
\begin{align*} H=&1-(1-F)^n(1-G)^m\\ I=& F^n\,G^m \end{align*} Although, we might not be able able to explicitly solve for $F$ and $G$, it is still possible to solve it numerically for $F(t)$ and $G(t)$ given the values of $H(t)$ and $I(t)$. Thus, you can identify $F$ and $G$ only using two order statistics: $Z_1$ and $Z_{m+n}$.