This answer gives an analytical approach for diagonal matrix elements.
First of all, since $r^k$ is spherically symmetric you can immediately integrate the angular parts:
\begin{equation}
\left< n' l' m' | r^k | n l m \right> = \left< n'l \right\| r^k \left\| nl \right> \delta_{l',l} \delta_{m',m},
\end{equation}
where $\left< n'l \right\| r^k \left\| nl \right>$ is called the reduced matrix element.
Diagonal matrix elements: expectation values of $r^k$
Diagonal matrix elements $\langle r^k \rangle = \left< nl \right\| r^k \left\| nl \right>$ can be calculated with a combination of the Hellmann-Feynman theorem and Kramers' relation without the need to perform any integral explicitly.
The Hellmann-Feynman theorem is given by
\begin{equation}
\frac{dE}{d\lambda} = \left< \psi(\lambda) \right| \frac{\partial H}{\partial \lambda} \left| \psi(\lambda) \right>,
\end{equation}
where $\left|\psi(\lambda) \right>$ is a normalized ket for all $\lambda$. In this case you take $\lambda = l$ (note that $n=n(l)$) for $k=-2$ and $\lambda = Z$ for $k=-1$. Here $l$ is the angular momentum quantum number and $Z$ the atomic number.
Kramers' relation is a recursion relation from which all other $\langle r^k \rangle$ for integer $k$ can be found. It can be derived (see link) from the radial Schroedinger equation of the hydrogen-like atom and is given by
\begin{equation}
\frac{k}{4} \left[ \left( 2l + 1 \right)^2 - k^2 \right] a_0^2 \langle r^{k-2} \rangle - Z \left( 2k + 1 \right) a_0 \langle r^{k-1} \rangle + Z^2 \frac{k+1}{n^2} \langle r^k \rangle = 0,
\end{equation}
where $a_0$ is the Bohr radius. Note that this relation by itself gives all $\langle r^k \rangle$ for $k>-2$ since $\langle r^0 \rangle = 1$.
Off-diagonal matrix elements
For the off-diagonal matrix elements, I looked in the 8th edition of Gradshteyn and Ryzhik (page 817) for an analytical expression of the relevant integrals, but none is given there. Likely, there are no analytical expressions for the off-diagonal matrix elements. This is also confirmed by the paper linked in the answer of Emilio Pisanty.