jmc

We re-write the given recursion as $$a_n=\frac{a_{n-1}^2}{b_{n-1}},b_n=\frac{b_{n-1}^2}{a_{n-1}}.$$Then $$a_n{b}_n=\frac{a_{n-1}^2}{b_n}\cdot \frac{b_{n-1}^2}{a_n}=a_{n-1}{b}_{n-1}.$$Solving for $a_{n-1}$ in $b_n=\frac{b_{n-1}^2}{a_{n-1}},$ we find $a_{n-1}=\frac{b_{n-1}^2}{b_n}.$ Then $a_n=\frac{b_n^2}{b_{n+1}}.$ Substituting into the equation above, we get $$\frac{b_n^2}{b_{n-1}}\cdot b_n=\frac{b_{n-1}^2}{b_{n+1}}\cdot b_{n-1}.$$Isolating $b_{n+1},$ we find $$b_{n+1}=\frac{b_{n-1}^4}{b_n^3}.$$We know that $b_0=3$ and $b_1=\frac{b_0^2}{a_0}=\frac{9}{2}.$ Let $$b_n=\frac{3^{s_n}}{2^{t_n}}.$$Then $s_0=1,$ $s_1=2,$ $t_0=0,$ and $t_1=1.$ From the equation $b_{n+1}=\frac{b_{n-1}^4}{b_n^3},$ $$\frac{3^{s_{n+1}}}{2^{t_{n+1}}}=\frac{{\left(\frac{3^{s_n}}{2^{t_n}}\right)}^4}{{\left(\frac{3^{s_{n-1}}}{2^{t_{n-1}}}\right)}^3}=\frac{3^{4s_n-3s_{n-1}}}{2^{4t_n-3t_{n-1}}},$$so $s_{n+1}=4s_n-3s_{n-1}$ and $t_{n+1}=4t_n-3t_{n-1}.$ We can then use these equations to crank out the first few terms with a table:

$$\begin{array}{ccc}n& s_n& t_n\\ 0& 1& 0\\ 1& 2& 1\\ 2& 5& 4\\ 3& 14& 13\\ 4& 41& 40\\ 5& 122& 121\\ 6& 365& 364\\ 7& 1094& 1093\\ 8& 3281& 3280\end{array}$$Hence, $(m,n)=\boxed{(3281,3280)}.$