By no means is this going to be a full answer to this question. However I am going to post it because I believe that this approach can eventually lead to a full solution. Our plan would be to express the sum in question through hypergeometric sums and then use identities governing the later to obtain closed form solutions.
To be specific let us take $m=3$ and $n$ being an odd integer. Then by using elementary properties of Pochhammer symbols and factorials we have:
\begin{eqnarray}
(3 i)! &=& 3^{3 i} \prod\limits_{\xi=0}^2 (1-\frac{\xi}{3})^{(i)}\\
(2 i-2n-1)! &=& 2^{2(i-n)-1} \cdot \left( \frac{1}{2} - n\right)^{(i)} \cdot \frac{(-n-1/2)!}{(-1/2)!} \cdot 1^{(i)} \cdot \frac{1}{\prod\limits_{\xi=0}^n(i-\xi) }\\
(i+2n+1)!&=& 1^{(i)} \cdot \prod\limits_{\xi=1}^{2 n+1} (i+\xi)
\end{eqnarray}
Therefore the sum in question reads:
\begin{eqnarray}
{\mathfrak S}^{(3)}_{2 n+1}(x) = \frac{(-1/2)!}{(-n-1/2)!} \sum\limits_{i=0}^\infty \frac{(2/3)^{(i)} \cdot (1/3)^{(i)} \cdot }{(1/2-n)^{(i)} } \cdot \frac{3^{3 i} x^i}{2^{2(i-n)-1} i!} \cdot \frac{\prod\limits_{\xi=0}^n (i-\xi)}{\prod\limits_{\xi=1}^{2n+1}(i+\xi)}
\end{eqnarray}
Now we are using the usual "tricks". We have:
\begin{eqnarray}
\prod\limits_{\xi=0}^n (i-\xi) &=& \left. \frac{d^{n+1}}{d \theta^{n+1}} \theta^i \right|_{\theta=1} \\
\frac{1}{\prod\limits_{\xi=1}^{2n+1}(i+\xi)} &=& \int\limits_0^1 \frac{(1-\xi)^{2 n}}{(2 n)!} \xi^i d\xi \\
\int\limits_0^1 \frac{(1-\xi)^{2 n}}{(2 n)!} \cdot \xi^{n+1} \cdot F_{2,1} \left[ \begin{array}{rr} a & b \\ c \end{array}; \xi x\right] d\xi &=& \frac{(c-1)_{(3n+2)}}{(a-1)_{(3n+2)} (b-1)_{(3n+2)}} \cdot \frac{d^{n+1}}{d x^{n+1}} \cdot \frac{1}{x^{2 n+1}} \left\{ F_{2,1} \left[ \begin{array}{rr} a-3n-2 & b-3n-2 \\c-3n-2\end{array};x\right] - \sum\limits_{l=0}^{3 n+1} \frac{(a-3n-2)^{(l)} (b-3n-2)^{(l)}}{(c-3n-2)^{(l)}} \frac{x^l}{l!}\right\}
\end{eqnarray}
Inserting those into the equation before them gives the following:
\begin{eqnarray}
&&{\mathfrak S}^{(3)}_{2 n+1}(x) = \frac{1}{2} \frac{(-1/2)!}{(-n-1/2)!} \cdot \frac{(2/3)^{(n+1)} (1/3)^{(n+1)} }{(1/2-n)^{(n+1)} } \cdot (27 x)^{n+1} \cdot \\
&&\int\limits_0^1 \frac{(1-\xi)^{2 n}}{(2n)!} \cdot \xi^{n+1} F_{2,1}\left[ \begin{array}{rr} 5/3+n & 4/3+n \\ 3/2 \end{array}; \frac{27}{4} x \xi\right] d\xi \\
&&=\frac{1}{2} \frac{(-1/2)!}{(-n-1/2)!} \cdot \frac{(2/3)^{(n+1)} (1/3)^{(n+1)} (-1/2-3n)^{(3n+2)}}{(1/2-n)^{(n+1)} (-1/3-2n)^{(3n+2)} (-2/3-2 n)^{(3n+2)} } \cdot (27 x)^{n+1} \cdot \\
&&\sum\limits_{p=0}^{n+1} \binom{n+1}{p} \frac{(-1)^{n+1-p}}{(27/4 x)^{3n+2-p}} \frac{(3n+1-p)!}{(2n)!} \left( \frac{(-1/3-2 n)^{(p)} (-2/3-2 n)^{(p)}}{(-1/2-3 n)^{(p)}} F_{2,1}\left[ \begin{array}{rr} -1/3-2n+p& -2/3-2 n+p \\ -1/2-3 n+p\end{array};\frac{27}{4} x\right]-\sum\limits_{l=p}^{3n+1} \frac{(-1/3-2 n)^{(l)} (-2/3-2 n)^{(l)}}{(-1/2-3 n)^{(l)}} \cdot \frac{(27/4 x)^{l-p}}{(l-p)!}\right)
\end{eqnarray}
Herewith we are done,i.e we have accomplished our goal. There are only hypergeometric functions in the right hand side. Now we will demonstrate that we can actually evaluate those quantities in closed form.
Let us take $n=0$ and $p=1$ first. Then we have:
\begin{eqnarray}
&&F_{2,1}\left[ \begin{array}{rr} -1/3-2n+p& -2/3-2 n+p \\ -1/2-3 n+p\end{array};\frac{27}{4} x\right] = F_{2,1}\left[ \begin{array}{rr} 2/3& 1/3 \\ 1/2\end{array};\frac{27}{4} x\right] =\\
&&\left(1- \frac{27}{4} x\right)^{-1/2} \cdot F_{2,1} \left[ \begin{array}{rr} 1/6 & -1/6 \\ 1/2\end{array};\frac{27}{4} x\right] =\\
&& \left(1- \frac{27}{4} x\right)^{-1/2} \cdot \cos \left( \frac{1}{6} \cdot \arccos\left( 1-\frac{27}{2} x\right)\right)=\\
&&\frac{\cos\left( \frac{1}{3} \cdot \arcsin \left( \frac{3 \sqrt{3}}{2} \sqrt{x}\right)\right)}{\sqrt{1-\frac{27}{4} x}}
\end{eqnarray}
where in the first line we used the Euler transformation of the hypergeometric sum and in the second line we used https://en.wikipedia.org/wiki/Hypergeometric_function#Other_points .
Now let us take $n=0$ and $p=0$. Then we have:
\begin{eqnarray}
&&F_{2,1}\left[ \begin{array}{rr} -1/3-2n+p& -2/3-2 n+p \\ -1/2-3 n+p\end{array};\frac{27}{4} x\right] = F_{2,1}\left[ \begin{array}{rr} -1/3& -2/3 \\ -1/2\end{array};\frac{27}{4} x\right] =\\
&&1- 3 \int\limits_0^1 F_{2,1}\left[ \begin{array}{rr} 2/3& 1/3 \\ 1/2\end{array};\frac{27}{4} \xi\right] d\xi = 1- 3 \int\limits_0^x \frac{\cos \left( \frac{1}{3} \cdot \arcsin\left( \frac{3 \sqrt{3}}{2} \sqrt{\xi}\right)\right)}{\sqrt{1- \frac{27}{4} \xi}}d\xi =\\
&& 1-4/9 \int\limits_0^{27/4 x} \frac{\cos\left(\frac{1}{3} \arcsin\left( \sqrt{\xi} \right)\right)}{\sqrt{1-\xi}}d\xi = \\
&&
\frac{1}{2} \left( \sqrt{4-27 x} \cos\left( \frac{1}{3} \arcsin\left( \frac{3 \sqrt{3} \sqrt{x}}{2}\right)\right) + \sqrt{3 x} \sin\left( \frac{1}{3} \arcsin\left( \frac{3 \sqrt{3} \sqrt{x}}{2}\right)\right)\right)
\end{eqnarray}
This concludes the case of $n=1$. We believe that this approach can be generalized to arbitrary values of $n$.
