Partitioning set $[n] = \{1, \dots, n\}$ into subsets of size at most $m$ such that all sums are powers of $m$

Given integer $n > 0$ and $m > 1$, we call a partition of set $[n] = \{1, \dots, n\}$ {\em $m$-good} if each of the partitioning sets is of size at most $m$ and the sum of numbers in it is a power of $m$, that is, $m^t$ for some $t \geq 0$. It is easily seen that a unique 2-good partition exists for each $n$ and, in contrast, for each fixed $m>3$, for infinitely many $n$, no $m$-good partition exists. Case $m=3$ is more difficult. We conjecture that 3-good partitions exist for each $n$ and prove that a minimal counter-example, if any, is at least 101 and must belong to the set {\centering $N_o = \{n \equiv 2 \pmod 3\} \cap \{3^t+1 < n < (3^{t+1}+1)/2 \mid t \geq 4\}.$ } For this case we provide some partial results. We also show that a 3-good partition is unique if $n \in N_u = \{1, 2, 3, 4, 3^t-4, 3^t-2, 3^t-1, 3^t, 3^t+1, 3^t+2, 3^t+3 \mid t > 1\}$ and conjecture that the inverse holds too.