- Question: $ \mbox{How can we evaluate}\quad \sum_{n \geq 0}\left[{4^{n} \over \left(\, 2n + 1\,\right) \binom{2n}{n}}\right]^{2}{1 \over n + k}\quad \mbox{for general $k$ ?.} $
General methodology will be enough, but a closed-form is more preferable (if exists). Note that the previous problem, i.e. expressing binomial series in terms of MZVs, is solved via an alternative method (by user @pisco), so I simplified the question. For his method see here.
The problem is equivalent to finding an explicit form for a $\phantom{}_4 F_3$ with half-integer parameters, since due to Rodrigues' formula and Euler's Beta function
$$\small \int_{0}^{1}\!\!\!P_n(2x-1)\sum_{m\geq 0}\left(\frac{4^m}{(2m+1)\binom{2m}{m}}\right)^2 x^m\,dx=\!\!\int_{0}^{1}\!\!\sum_{m\geq n}\left(\frac{4^m}{(2m+1)\binom{2m}{m}}\right)^2\binom{m}{n}x^{m}(1-x)^n\,dx $$ equals $$ \frac{16^n}{(2n+1)^3\binom{2n}{n}^3}\cdot\phantom{}_4 F_3\left(n+1,n+1,n+1,n+1;n+\tfrac{3}{2},n+\tfrac{3}{2},2n+2;1\right).$$ Not surprising since $\phantom{}_4 F_3(1^{(4)};3/2^{(2)},2;x)$ essentially is the primitive of $\phantom{}_3 F_2(1^{(3)};3/2^{(2)};x)$.
Let us see if we manage to crack the case $n=0$: $$ \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+1}=\sum_{n\geq 0}\frac{4^n}{(2n+1)(n+1)\binom{2n}{n}}\int_{0}^{\pi/2}\left(\sin\theta\right)^{2n+1}\,d\theta $$ due to the Maclaurin series of $\arcsin(x)^2$ equals $$ \int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = 2\pi\, C-\frac{7}{2}\zeta(3)$$ and I guess this method can be applied to other values of $n$, too.
For instance, for $n=1$ we have to find $$ \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{16(n+1)^3}{(n+2)(n+3)(2n+3)^2} $$ which by partial fractions decomposition boils down to evaluating $$\small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+A+1},\quad \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{2n+2B+1},\quad \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{(2n+2B+1)^2} $$ for specific values of $A,B\in\mathbb{N}$. The situation is the same for $n>1$.
A small collection of relevant identities: $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+2} = -\frac{1}{4}+\frac{\pi}{4}+\frac{\pi C}{2}-\frac{7\zeta(3)}{8} $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+3} = -\frac{11}{64}+\frac{13\pi}{64}+\frac{9\pi C}{32}-\frac{63\zeta(3)}{128} $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{2n+1} = -\pi\,C+\frac{7}{2}\zeta(3) $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{2n+3} = -1+\frac{\pi}{2} $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{(2n+3)^2} = -3+\pi. $$
It is relevant to point out that $$ \frac{4^n}{(2n+1)\binom{2n}{n}}=\frac{2n+3}{2n+2}\cdot\frac{4^{n+1}}{(2n+3)\binom{2n+2}{n+1}} $$ so reindexing (together with the identities in this answer) is extremely useful for dealing with series of the last kind.
There is also this nice result that John Campbell, Marco Cantarini and I proved through fractional operators: if $f\in(C^{\omega}\cap L^2)(0,1)$ is such that $$ f(x)=\sum_{n\geq 0}a_n x^n = \sum_{m\geq 0} b_m P_m(2x-1) $$ then $$ \sum_{n\geq 0}\frac{a_n}{(2n+1)^2\left[\frac{1}{4^n}\binom{2n}{n}\right]^2} = \sum_{m\geq 0}\frac{(-1)^m b_m}{(2m+1)^2}.$$
Partial fraction decomposition then shows that for any $k\in\mathbb{Z}^+$ your series is easily converted into a linear combination with rational coefficients of $1,\pi,\pi C$ and $\zeta(3)$ through the FL-expansion of $\frac{1}{x^k}\left(-\log(1-x)-\sum_{s=1}^{k-1}\frac{x^k}{k}\right)$, which can be derived from $$ -\log(1-x)=1+\sum_{m\geq 1}(-1)^m\left(\frac{1}{m}+\frac{1}{m+1}\right)P_m(2x-1) $$ and the method previously outlined here.