<s>[INST] Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square. Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.
I found these pairs work by mod 3 and mode 4:
Case 1, $2^m \ge 3^n$, $(0,0), (2,1), (1,0)$
Case 2, $2^m \le 3^n$, $(0,0), (1,1), (1,3), (3, 2), (5, 4)$
How to prove there's no others?
I also found this post: Pairs forming perfect square. There's good discussions. Do we have any complete solution?
 [/INST] This is a partial solution concerning $m\ne 1$.
First we consider the cases $m\ge 2$.
For case 1 ($2^m \ge 3^n$), write $2^m = 3^n + k^2$.
Taking modulo $3$, we have $(-1)^m \equiv k^2 \pmod 3$. This gives $m$ is even.
Write $m = 2s$, then $3^n = 2^m-k^2 = (2^s-k)(2^s+k)$, and both $2^s \pm k$ are powers of $3$.
Summing them up, however, gives $2^{s+1}$ which is not a multiple of $3$. Hence $2^s-k = 1$.
$2^s+k = 2^{s+1}-1$ is a power of $3$. [This gives $s=1, m=2, n=1$.]

For case 2 ($2^m \le 3^n$), write $3^n = 2^m + k^2$.
Taking modulo $4$, we have $(-1)^n\equiv k^2 \pmod 4$. This gives $n$ is even.
Write $n=2t$, then $2^m = 3^n-k^2 = (3^t-k)(3^t+k)$, and both $3^t \pm k$ are powers of $2$.
If we write $3^t-k = 2^x, 3^t+k =2^y$, we have $2^x+2^y = 2^x(1+2^{y-x})$.
Summing them up, however, gives $2 \times 3^t$. Since $3^t$ is odd, we must have $2^x=2$ and $1+2^{y-1} = 3^t$.
[This equation has $2$ solutions: $y=2, t=1$ and $y=4, t=2$.]

Both cases above end with solving an equation of the form $1+a^b= c^d$, which by Catalan's Conjecture/Mihăilescu's theorem only has the solution $a=2,b=3,c=3,d=2$, for integers $a,b,c,d >1$. However there are elementary proofs for this special case $a,c \in \{2,3\}$, which I choose to omit (unless requested)

For the case $m=0$, write $3^n = 1+k^2$. By modulo $3$, no solutions exist for $n \ge 1$. This forces $n=0$.
For the case $m=1$, write $3^n = 2+k^2$. By modulo $3$ we have $k^2 \equiv 1\pmod 3$, and by modulo $4$ we have $(-1)^n \equiv 2+k^2$, forcing $n$ to be odd and $k^2\equiv 1 \pmod 4$. $k$ must be an odd number not divisible by $3$, but as of right now I am not sure how to proceed.
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