Let $f$ be define by the power series \begin{align} f(x)=\sum_{n=0}^\infty a_n x^n. \end{align} where $f(1)=1$.
Moreover, let $f$ coefficient of power series satisfy \begin{align} \sum_{n=0}^\infty (n+1) a_{n+1} x^n = \sum_{n=0}^\infty ( c_k n^k+c_{k-1} n^{k-1}+ \cdots +c_0 ) a_n x^n \end{align} for some fixed and given $c_k,...,c_0$. We have to solve for $f$ or find $a_k$'s.
My approach is to solve it by trying to re-write the above as differential equation in terms of $f$. For example, the first term can be written as \begin{align} \sum_{n=0}^\infty (n+1) a_{n+1} x^n= \sum_{n=1}^\infty n a_{n} x^{n-1}=f'(x) \end{align}
So we have that \begin{align} f'(x)= \sum_{n=0}^\infty ( c_k n^k+c_{k-1} n^{k-1}+\cdots+c_0 ) a_n x^n \end{align}
Now how to re-write $\sum_{n=0}^\infty ( c_k n^k+c_{k-1} n^{k-1}+\cdots+c_0 ) a_n x^n$ in terms of derivatives of $f$?
Note that $$ D_x^j(x^jf(x))=\sum_{n\ge0}a_n n^{(j+1)}x^n $$ with $n^{(\,\cdot\,)}$ the rising factorial. Thus, $$\begin{aligned} \sum_{n\ge0} c_0 a_n x^n&=c_0f(x)\\ \sum_{n\ge0} c_1 n a_nx^n&=c_1\left[(xf(x))'-f(x)\right]\\ \sum_{n\ge0} c_2 n^2a_nx^n&=c_2\left[(x^2f(x))''-3\left[(xf(x))'-f(x)\right]-2f(x)\right]\\ &\cdots \end{aligned} $$
Therefore, your function indeed satisfies a linear differential equation of order $k$. Writing it down explicitly for arbitrary $\{c_i\}$ is left as an exercise to the reader. Hint: think of the Stirling numbers of the second kind:
Note also that one may write $f(x)$ as a $_kF_0$ hypergeometric function whose parameters are essentially the roots of $c_kn^n+\cdots+c_0=0$. Again, for arbitrary $\{c_i\}$ this is as explicit as it gets.