Why do expressions of the form $\sum\limits^\infty_{n=1} \frac{n^k}{3^n}$ sum 'nicely'?

The nice identity related to such sums is $$ \sum_{n\ge 0} \binom{n+k}{k} x^k = \frac1{(1-x)^{k+1}}. $$ You can prove this by induction, or by generating functions, or by the generalized binomial theorem.

The values of your sums aren't going to be as nice, but we can express any monomial $n^k$ as a linear combination of $\binom n0, \binom{n+1}{1}, \dots, \binom{n+k}{k}$ because these are a basis for polynomials of degree up to $k$. As a result, we know that $$ \sum_{n \ge 0} n^k x^k = \frac{a_0}{1-x} + \frac{a_1}{(1-x)^2} + \dots + \frac{a_k}{(1-x)^{k+1}} $$ for some coefficients $a_0, \dots, a_k$, which we can put over a common denominator of $(1-x)^{k+1}$.

In your case, the denominator for $x = \frac12$ simplifies to a factor of $2^{k+1}$, the denominator for $x = \frac13$ simplifies to a factor of $(\frac32)^{k+1}$, and they get progressively worse from there.

Another way. Note that $$ n^k=\sum_{m=0}^kS(k,m)n^{\underline{m}} $$ where $S(k,m)$ is the stirling number of the second kind and $n^{\underline{m}}=n(n-1)\dotsb(n-m+1)$ is the falling factorial of length $m$. In effect we have simply changed bases to the more convenient basis of falling factorials which are a basis for polynomials of degree at most $k$. Now $$ \sum_{n=0}^\infty n^k x^n=\sum_{n=0}^\infty\left(\sum_{m=0}^kS(k,m)n^{\underline{m}}\right)x^n =\sum_{m=0}^kS(k,m)\left(\sum_{n=0}^\infty n^{\underline{m}}x^n \right)\quad (|x|<1)\tag{1} $$ But it is easy to get the generating function of $n^{\underline{m}}$ by differentiation of the geometric series. Indeed, $$ \sum_{n=0}^\infty n^{\underline{m}}x^n=x^m\left[\frac{d^m}{dx^m}(1-x)^{-1}\right]=\frac{m!x^m}{(1-x)^{m+1}}\quad (|x|<1) $$ whence $$ \sum_{n=0}^\infty n^k x^n=\sum_{m=0}^k \left[m!S(k,m)\frac{x^m}{(1-x)^{m+1}}\right].\quad(|x|<1). $$ For example put $x=1/2$ to get that $$ \sum_{n=0}^\infty n^k 2^{-n}= 2\sum_{m=0}^k m!S(k,m). $$ and put $x=1/3$ to get that $$ \sum_{n=0}^\infty n^k 3^{-n}= 3\sum_{m=0}^k m!S(k,m)\frac{1}{2^{m+1}}. $$ and in general put $x=1/r$ where $|x|<1$, to get that $$ \sum_{n=0}^\infty n^k r^{-n}=r\sum_{m=0}^km!S(k,m)\frac{1}{(r-1)^{m+1}}. $$