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 $t\ge 0$. 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 $N_o=\{n\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)\}\cap \{3^t+1<n<(3^{t+1}+1)\mathrm{/}2\mid t\ge 4\}\mathrm{.}$ } 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.