Those things are not really very hard only quite tedious since it takes a lot of time to solve them even when using Mathematica, We want to compute;
\begin{equation}
lhs_{k,m} := F_{2,1}\left[ \begin{array}{cc} 2k & k+m \\ 2k+2 m \end{array}; 1- \frac{1}{x} \right]
\end{equation}
Firstly take $m=0$ then:
\begin{eqnarray}
lhs_{k,0}&=&\sum\limits_{p=0}^\infty (k+m)^{(p)} \frac{(1-\frac{1}{x})^p}{p!}\\
&=& \sum\limits_{p=0}^\infty \binom{k+m+p-1}{p} (1-\frac{1}{x})^p \\
&=& \sum\limits_{p=0}^\infty \binom{-k-m}{p} (\frac{1}{x}-1)^p = (1+\frac{1}{x} - 1)^{-k-m}=x^{k+m}
\end{eqnarray}
Now we take $m=1$.
\begin{eqnarray}
lhs_{k,1}&=& \sum\limits_{p=0}^\infty \frac{(2k)(2k+1)}{(2k+p)(2k+p+1)} \cdot (k+m)^{(p)} \frac{(1-\frac{1}{x})^p}{p!}\\
&=& (2k)(2k+1) \sum\limits_{p=0}^\infty \left( \frac{1}{2k+p} - \frac{1}{2k+p+1} \right) \binom{-k-1}{p} (\frac{1}{x} - 1)^p \\
&=& (2k)^{(2)} \int\limits_0^1 \left( t^{2k-1} - t^{2 k} \right) (1+t (\frac{1}{x} - 1))^{-k-1} dt \\
&=& (2k)^{(2)} \frac{x^{2 k}}{(1-x)^{2k+1}} \int\limits_1^{1/x} \frac{(u-1)^{2k-1}}{u^{k+1}} (1-x u) du \\
&=&(2k)^{(2)} \frac{x^{2 k}}{(1-x)^{2k+1}} \sum\limits_{p=0}^{2k-1} \binom{2k-1}{p} (-1)^{p+1}\cdot \\
&&
\left(
\frac{(\frac{1}{x})^{p-k} - 1}{p-k} 1_{k\neq p} + \log(1/x) 1_{k=p} - x \frac{(\frac{1}{x})^{p-k+1} - 1}{p-k+1} 1_{k \neq p+1} - x \log(1/x) 1_{k=p+1}
\right)
\end{eqnarray}
From the above it is not hard to see that those calculations can be carried through to higher values of $m$ and then the generic result can be proven by induction. The result reads:
\begin{eqnarray}
lhs_{k,m}= \frac{(2k)^{(2m)}}{(2m-1)!} \sum\limits_{q=0}^{2m-1} \binom{2m-1}{q} (-1)^q \frac{x^{2k+q}}{(1-x)^{2k+q}} \sum\limits_{p=0}^{2k+q-1} \binom{2k+q-1}{p} (-1)^{q-1-p}
\left(
\frac{(\frac{1}{x})^{p-k-m+1} -1}{p-k-m+1} 1_{p-k-m \neq 1} + \log(\frac{1}{x}) 1_{p-k-m=1}
\right)
\end{eqnarray}
for $k=1,2,\cdots$ and $m=1,2,\cdots$ and $lhs_{0,m}=1$ and $lhs_{k,0}=x^k$.
In[18]:= mmax = 5;
mat = Table[
Hypergeometric2F1[2 k, k + m, 2 k + 2 m, 1 - 1/x], {k, 0,
mmax}, {m, 0, mmax}];
tmp = Simplify[
Table[Pochhammer[2 k, 2 m]/(2 m - 1)! Sum[
Binomial[2 m - 1, q] (-1)^q x^(2 k + q)/(1 - x)^(2 k + q) Sum[
Binomial[2 k + q - 1, p] (-1)^(q - 1 - p) If[
p - k - m != -1, (1/x^(p - k - m + 1) - 1)/(p - k - m + 1),
Log[1/x]], {p, 0, 2 k + q - 1}], {q, 0, 2 m - 1}], {k, 0,
mmax}, {m, 0, mmax}] - mat, Assumptions -> x > 1];
MatrixForm[tmp]
