You can solve the problem in three cases:
Case 1: $m\ge2$ and $2^m>3^n$.
Here, looking modulos $3$ and $4$ demands that $m$ is even and $n$ is odd. The problem then reduces to the form
$$(2^{m/2})^2=x^2+3(3^{(n-1)/2})^2\,~\,~\,~\,~ (1)$$
for some integer $x$; thus we are looking at solutions to the Diophantine equation
$$z^2=x^2+3y^2\,.$$
Checking modulo $8$ shows that the only possibility for both $x,y$ to be odd is when $z^2\equiv 4\mod 8$, which thus demands that the only solutions to Eq (1) must be $$(m,n)=(2,1)\,.$$
Case 2: $m\ge2$ and $2^m<3^n$.
Here also, looking modulos $3$ and $4$ rather demands that $m$ is odd and $n$ is even. Now, the problem reduces to the form
$$(3^{n/2})^2=x^2+2(2^{(m-1)/2})^2\,~\,~\,~\,~ (2)$$
for some integer $x$; thus, this time, we are looking at solutions to the Diophantine equation
$$z^2=x^2+2y^2\,.$$
The general solution to this equation (in the manner similar to deriving Pythagorean Triples) is given by $$z=a^2+2b^2\,,~\,~\,~\,x=a^2-2b^2\,,~\,~\,~\,~y=2ab$$ for some integers $a,b$. Comparing to Eq (2) forces $a=\pm 1$ and $b=\pm 2^{(m-3)/2}$, which implies that $3^{n/2}=z=(\pm 1)^2+2(\pm 2^{(m-3)/2})^2= 1+2^{m-2}$——that is, $3^{n/2}-2^{m-2}=1$——which has the only solutions $$(m,n)\in\{(3,2),(5,4)\}$$ thanks Catalan-Mihailescu’s Theorem.
Case: $m\in\{0,1\}$.
For $n=0$, we have the obvious solutions $$(m,n)\in\{0,0),(1,0)\}\,.$$ Hence suppose $n\ge 1$. Here, we have $3^n-2^m=y^2$ for some integer $y$, which can be rewritten as
$$y^2=3^{e_n}\left(3^{(n-e_n)/3)}\right)^3-2^m\,,$$ where $e_n\in\{0,1,2\}$ and $n\equiv e_n\mod 3$,. Multiplying through by $3^{2e_n}$, we see that we are actually looking at integer solutions to the Mordell elliptic curve $$y^2=x^3-2^m\cdot 3^{2e_n}\,,$$ and a solution to your problem corresponds to $x=3^{(n+2e_n)/3}$. Mordell (1920) proved that there are only finitely many integral solutions to such elliptic curves (you can find the integer solutions listed here https://web.archive.org/web/20110618033146/http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL-); in particular, for $m=1$ and $e_n=0$, it is a theorem or Fermat that the only integral solutions to the elliptic are $(x,y)\in\{3,\pm 5\}$, which thus gives the only solution to your problem as $$(m,n)=(1,3)\,.$$ When $m=e_n=0$, the only integral solution is $(x,y)=(1,0)$, which doesn’t correspond to a solution to your problem.
For $e_n=1$, using the linked address above, the only integral solutions are no solutions for $m=0$ but $(x,y)=(3,\pm 3)$ for $m=1$, which forces the only solution to your problem in this case $$(m,n)=(1,1)\,.$$ Finally, for $e_n=2$, there are no integral solutions to the elliptic curve for $m=1$, but for $m=0$, the only solution is $(x,y)=(13,46)$, which does not correspond to a solution to your problem.