Here is another variation of the theme. It's not the shortest one, but it could be helpful for similar cases and it provides a slightly different representation of the solution.
Since we often have to cope with series of the form
\begin{align*}
G(x)=\frac{p+qx}{ax^2+bx+c}
\end{align*}
we derive a general representation by a series expansion at $x=0$. Based upon this result we obtain a solution for $F(x)$ and $F_m(x)$.
Using partial fraction decomposition of $G(x)$ and denoting the zeros of $ax^2+bx+c$ with $x_0,x_1$ we obtain
\begin{align*}
G(x)&=\frac{p+qx}{ax^2+bx+c}\\
&=\frac{1}{a}\frac{p+qx_0}{x_0-x_1}\frac{1}{x-x_0}-\frac{1}{a}\frac{p+qx_1}{x_0-x_1}\frac{1}{x-x_1}\tag{1}\\
&=\frac{1}{ax_0}\frac{p+qx_0}{x_1-x_0}\frac{1}{1-\frac{x}{x_0}}
-\frac{1}{ax_1}\frac{p+qx_1}{x_1-x_0}\frac{1}{1-\frac{x}{x_1}}\\
&=\frac{1}{ax_0}\frac{p+qx_0}{x_1-x_0}\sum_{k=0}^{\infty}\left(\frac{x}{x_0}\right)^k
-\frac{1}{ax_1}\frac{p+qx_1}{x_1-x_0}\sum_{k=0}^{\infty}\left(\frac{x}{x_1}\right)^k\tag{2}\\
&=\frac{x_1}{c}\frac{p+qx_0}{x_1-x_0}\sum_{k=0}^{\infty}\left(\frac{a}{c}x_1\right)^kx^k
-\frac{x_0}{c}\frac{p+qx_1}{x_1-x_0}\sum_{k=0}^{\infty}\left(\frac{a}{c}x_0\right)^kx^k\tag{3}
\end{align*}
Comment:
In (1) we use partial fraction decomposition and $$ax^2+bx+c=a(x-x_0)(x-x_1)=a(x^2-(x_0+x_1)x+x_0x_1)$$
In (2) we represent the fractions as geometric series at $x=0$.
In (3) we use the relationship $x_0x_1=\frac{c}{a}$
In the following we use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of the series in (3)
\begin{align*}
[x^n]G(x)&=\frac{p+qx_0}{x_1-x_0}a^n\left(\frac{x_1}{c}\right)^{n+1}
-\frac{p+qx_1}{x_1-x_0}a^n\left(\frac{x_0}{c}\right)^{n+1}\\
&=\frac{a^n}{c^{n+1}}\frac{1}{x_1-x_0}\left[(p+qx_0)x_1^{n+1}-(p+qx_1)x_0^{n+1}\right]
\end{align*}
Now it's time to harvest. With
\begin{align*}
F(x)&=\frac{p+qx}{ax^2+bx+c}=\frac{1+x}{-2x^2-2x+1}
\end{align*}
we obtain
\begin{align*}
x_{0,1}&=\frac{1}{2a}\left(-b\pm\sqrt{b^2-4ac}\right)=-\frac{1}{2}(1\pm\sqrt{3})\\
x_1-x_0&=\sqrt{3}\\
p+qx_0&=\frac{1}{2}(1-\sqrt{3})\\
p+qx_1&=\frac{1}{2}(1+\sqrt{3})
\end{align*}
and we derive the coefficient $[x^n]F(x)$ as
\begin{align*}
[x^n]F(x)&=(-2)^n\frac{1}{\sqrt{3}}\left[\frac{1}{2}(1-\sqrt{3})\left(-\frac{1}{2}\right)^{n+1}(1-\sqrt{3})^{n+1}\right.\\
&\qquad\qquad\qquad\left.-\frac{1}{2}(1+\sqrt{3})\left(-\frac{1}{2}\right)^{n+1}(1+\sqrt{3})^{n+1}\right]\\
&=\frac{1}{4\sqrt{3}}\left[(1+\sqrt{3})^{n+2}-(1-\sqrt{3})^{n+2}\right]\\
&=\frac{1}{4\sqrt{3}}\sum_{j}2\binom{n+2}{2j+1}\left(\sqrt{3}\right)^{2j+1}\\
&=\frac{1}{2}\sum_{j}\binom{n+2}{2j+1}3^j\tag{4}
\end{align*}
Note, that from (4) and OPs representation we obtain the identity
\begin{align*}
\frac{1}{2}\sum_{j}\binom{n+2}{2j+1}3^j=\sum_{j}\left[\binom{n+1}{2j+1}+\binom{n}{2j+1}\right]3^j
\end{align*}
$$ $$
Similarly we obtain from the generating function
\begin{align*}
F_m(x)=\frac{1+x}{-mx^2-mx+1}
\end{align*}
the coefficient $[x^n]$ of $F_m$:
\begin{align*}
[x^n]F_m(x)&=\frac{\left(\frac{m}{2}\right)^n}{4\sqrt{1+\frac{4}{m}}}
\left[\left(1+\sqrt{1+\frac{4}{m}}\right)^{n+2}-\left(1-\sqrt{1+\frac{4}{m}}\right)^{n+2}\right]\\
&=\frac{1}{2}\left(\frac{m}{2}\right)^n\sum_{j}\binom{n+2}{2j+1}\left(1+\frac{4}{m}\right)^j
\end{align*}
Note: With the help of Wolfram Alpha we find
\begin{align*}
F(x)=\frac{1+x}{-2x^2-2x+1}=1+3x+8x^2+22x^3+60x^4+164x^5+\mathcal{O}(x^6)
\end{align*}
Regarding a comment, note the generating function of Jack D'Aurizio is
\begin{align*}
H(x)&=\frac{3x+2x^2}{-2x^2-2x+1}\\
&=F(x)-1\\
&=3x+8x^2+22x^3+60x^4+164x^5+\mathcal{O}(x^6)
\end{align*}
