Yet another family of hypergeometric sums that has a closed form solution. Let $m \ge 2$ and $j\ge 0$ be integers. Now, let $0 < z < \frac{(m-1)^{m-1}}{m^m}$ be a real number. Consider a following sum:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) := \sum\limits_{i=0}^\infty \binom{m \cdot i + j}{i}\cdot z^i
\end{equation}
By using both my answer to Closed form solutions for a family of hypergeometric sums. and by generalizing the approach from my answer to About the identity $\sum\limits_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$  I have derived the following results:
\begin{eqnarray}
{\mathfrak S}^{(m,0)}(z) &=& \frac{x \left(1-z x^{m-1}\right)}{1-m z x^{m-1}}\\
{\mathfrak S}^{(m,1)}(z) &=& \frac{m}{m-1} \cdot\frac{ x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-\frac{1}{m-1}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}\\
{\mathfrak S}^{(m,2)}(z) &=& \frac{m^2}{(m-1)^2}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}
-\frac{(m-2) }{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&
-\frac{1}{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}} \cdot \frac{((m-1) x+2) }{ x}\\
{\mathfrak S}^{(m,3)}(z) &=& \frac{m^3}{(m-1)^3}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-
\frac{(m-2)(2 m-3)}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&-\frac{m-3}{(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}}\cdot \frac{((m-1) x+2)}{x}+\\
&&-\frac{1}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{3}{m-1}}\cdot
\frac{2(m-1)^2 x^2+6(m-1)x+3(m+2)}{x^2}
\end{eqnarray}
Here $x:=x(z)$ is obtained in the following way. Out of the solutions of the equation:
\begin{equation}
1-x+z \cdot x^m=0
\end{equation}
we choose the one that is the closest to unity.
Now the question is how does the result look like for arbitrary $j \ge 2$.

Here I am using the results provided above by Markus Scheuer. Firstly I provide a closed form expression for the quantity $B_m(z)$. We have:
\begin{eqnarray}
B_m(z)&=& \sum\limits_{i=0}^\infty \binom{m\cdot i}{i} \cdot \underbrace{\frac{1}{(m-1) i+1}}_{\int\limits_0^1 \theta^{(m-1) i} d\theta} \cdot z^i\\
&=& \int\limits_0^1 \frac{x\cdot(1-z\theta^{m-1}\cdot x^{m-1})}{1-m z \theta^{m-1} \cdot x^{m-1}} d\theta \\
&\underbrace{=}_{z \theta^{m-1}=(\xi-1)/\xi^m}& \frac{1}{z^{1/(m-1)} \cdot(m-1)}\int\limits_0^{x-1} \frac{\xi^{1/(m-1)-1}}{(1+\xi)^{m/(m-1)}} d\xi\\
&=& \left( \frac{x-1}{z \cdot x}\right)^{1/(m-1)} = x
\end{eqnarray}
where $1-x+ z \cdot x^m=0$.
Therefore using the result quoted by Markus Scheuers'  we have:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) = \frac{(\frac{x-1}{z\cdot x})^{j/(m-1)}}{1-m+m \cdot (\frac{z \cdot x}{x-1})^{1/(m-1)}} = \frac{x^{j+1}}{(1-m) \cdot x+m}
\end{equation}
In particular if $m=1,2$ the result reads:
\begin{eqnarray}
\left\{ {\mathfrak S}^{(m,j)}\right\}_{m=1}^2 = \left\{ (\frac{1}{1-z})^{j+1}, \frac{(\frac{1-\sqrt{1-4 z}}{2 z})^j}{\sqrt{1-4 z}}\right\}
\end{eqnarray}
as it should be.
