Take $x \in \Bbb N$, $x \le9$ and $m \in \Bbb N$. Now we define a function $d_s(n): \Bbb N \to \Bbb N$ as the digit sum of $n$ in base $10$. Now let's say we have a lower bound $b_l$ and an upper bound $b_u$ (both implicitly as functions of $m$) so that $b_l \le d_s(x^m) \le b_u$. Find the values $b_l$ and $b_u$.
The only things i have come up with are $b_u \le 9 \lfloor m*log_{10}(x)+1 \rfloor$ since that is digit sum of an integer with the same number of digits, all of them having a value of $9$ and $b_l \ge 1$ for obvious reasons. I can't seem to find a way to approach either of these values other than the trivial remarks I made earlier. I used Mathematica to evaluate some values and there does seem to be a pattern.
Here's the rough outline of the argument I mentioned in the comments that $b_l\gt 3$ in all non-trivial cases: Suppose $d(x^m)=1$ for some $m$. Then this implies that $x^m=10^i$ (since the only numbers with $d(n)=1$ are powers of $10$); but then it must be the case that $5|x$ and therefore $x=5$ (since $x\leq 9$), but clearly powers of $5$ aren't powers of $10$ (for one thing, they're odd).
Now, if $d(x^m)=2$ then things break down into two cases: $x^m=2\cdot 10^i$ for some $i$, or $x^m=10^i+10^j$. The first case can be handled using the same argument that $5|x$; this also forces $j=0$ in the second case, so that we're left with $x^m=10^i+1$. Now, if this is ever to be the case then we must have $x$ odd (since $2\not\mid 10^i+1$); $x=3$ and $x=9$ are impossible since $10^i+1\not\equiv0\pmod 3$, and similarly $x=5$ is impossible since $10^i+1\not\equiv 0\pmod 5$, so the only possibility is $x=7$. But now if $10^i\equiv-1\pmod 7$ then $i\equiv 3\pmod 6$ (you should be able to show this easily) and in particular $i$ is odd — but this implies that $11|10^i+1$ (since then $10^i\equiv (-1)^i\equiv -1\pmod {11}$) and so it's impossible that $10^i+1\equiv 7^m$.
Finally, if $d(x^m)=3$ then $3|x^m$ and so $3|x$ — in which case, $3^m|x^m$ and in particular $9|x^m$ (except for the trivial case $m=1$). But of course, if $9|x^m$ then $9|d(x^m)$, giving the needed contradiction.