I did find a solution but did not prove uniqueness.  I'll present this as I worked it out, though somewhat verbose.

Each of (3m+4n) and (4m+3n) must be a power of 3, and the powers must add to 63.

n must be a multiple of 3, otherwise (3m+4n) would not be divisible by 3, let alone be a power of 3.

Similarly, m must be a multiple of 3.

Consider (3m+4n).  Look for (m,n) to make this a power of 3.  If m is a power of 3 and n=0 we have success, but then (4m+3n) will be an even number.  So m,n cannot be 0.

The smallest success where both m,n are multiples of 3 is (15,9): 45+36=81.  But then the other expression is 87.

Searching up to a few thousand, I can find no (m,n) pairs where both expressions are powers of 3.  I suspect there is no solution.

Suppose we find (4m+3n) which is a power of 3, say 3^x.  The value of the other expression is (3^x + m - n).  I expect (4m+3n) to be much larger than m or n, so it seems unlikely that (3^x + m - n) would differ enough from 3^x to be a different power of 3.  Of course if m=n, we have a brief success until realizing that then (4m+3n) = (4m+3m) = 7n and therefore not a power of 3.

For (3^x + m - n) to be 3^(x+1), 

m-n = 2*3^x

m = n + 2*3^x

Substituting into (3m+4n)=3^x

3(n + 2*3^x) + 4n = 3n + 4n + 6*3^x

but 7n+6*3^x cannot equal 3^x.  Rejected.

For (3^x + m - n) to be 3^(x-1), 

n-m = 2*3^(x-1)

n = m + 2*3^(x-1)

Substituting into (3m+4n)=3^(x-1)

3m + 4(m + 2*3^(x-1))

7m + 8*3^(x-1) = 3^(x-1)  Rejected.

Unless ...

I haven't considered negative values for one of (m, n). 

(45, -33): 3 and 81; powers 1+4 = 5

           [and note that (-33, 45) yields 81 and 3; so either order]

(135, -99): 9 and 243; powers 2+5 = 7

(405, -297):  27 and 729; powers 3+6 = 9

(1215, -891):  81 and 2187 ; powers 4+7 = 11

(15*3^a, -11*3^a): 3^a and 3^(a+3); sum powers = 2a+3

for 2a+3 = 63, a=30

(15*3^30, -11*3^30)

So we need the powers to total 63; powers 30+33

(m,n) = (15*3^30, -11*3^30)

(m,n) = (3088366981419735, -2264802453041139) in either order

So this counts as two solutions if you count (n,m) as distinct from (m,n) or one solution if you don't.

But I have not proved this is the only solution, so there could be more.