<s>[INST] Find the solutions to $\left\lfloor\left(\frac{5}{3} \right)^n\right\rfloor = 3^m$ 
Find all positive integer solutions to $\left\lfloor\left(\dfrac{5}{3}
\right)^n\right\rfloor = 3^m$.

Let $a_n = \left\lfloor\left(\dfrac{5}{3}
\right)^n\right\rfloor$. Then $$a_n = 1,2,4,7,12,21,35,59,99,165,275,459,765,1276,2126,3544,5907,9846,16410,\ldots.$$ Since a power of $3$ is odd, we only need to look at the odd terms of $a_n$: let these be $b_n$. Then $b_n = 1,7,21,35,59,99,165,275,459,765,5907,\ldots$. There don't seem to be powers of $3$ with a positive exponent in $b_n$ in the first few terms.
Using the Binomial Theorem, we have $$\left(\dfrac{5}{3}
\right)^n = \left(1+\dfrac{2}{3}\right)^n = 1+\dfrac{2}{3} \binom{n}{1}+\left(\dfrac{2}{3}\right)^2 \binom{n}{2}+\cdots+\left(\dfrac{2}{3}\right)^n\binom{n}{n}.$$ How can we continue from here?
 [/INST] If we write
$$
\left(\frac{5}{3}\right)^n = \left[ \left(\frac{5}{3} \right)^n \right] + \left\{ \left(\frac{5}{3}\right)^n \right\},
$$
then supposing that 
$$
\left[ \left(\frac{5}{3} \right)^n \right]=3^m
$$
implies the inequality
$$
0 < 5^n - 3^{n+m} < 3^n.
$$
Applying lower bounds for linear forms in (two complex) logarithms (say from Laurent-Mignotte-Nesterenko) to
$$
\Lambda = n \log 5 - (n+m) \log 3
$$
leads, after a wee bit of work, to the conclusion that $n \leq 1$. 
I'm not sure if there's an elementary way to prove this. I can't see one off the top of my head.
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