$\newcommand\bi\binom\newcommand{\tr}{\tilde r}$In view of Rob Pratt's comment, it is enough to show that
\begin{equation*}
L_m:=A_m-B_m\overset{\text{(?)}}\ge R_m \tag{10}\label{10}
\end{equation*}
for integers $m,n\ge1$, where
\begin{equation*}
A_m:=A_{m,n}:=\bi{m+4n}{m+n},\quad B_m:=B_{m,n}:=\bi{m+3n}{m+n},\quad
R_m:=R_{m,n}:=\bi{4m+4n}{n}.
\end{equation*}

Let
\begin{equation*}
r_m:=\frac{B_m}{A_m},\quad \rho_m:=\frac{R_m}{A_m},
\end{equation*}
so that $L_m=A_m(1-r_m)$ and $R_m=A_m\rho_m$. So, it is enough to show that
\begin{equation*}
r_m+\rho_m\overset{\text{(?)}} \le1. \tag{20}\label{20}
\end{equation*}

For any integers $a,b,k$ such that $0\le k\le a\le b\ne0$,
\begin{equation*}
\frac{\bi ak}{\bi bk}=\frac ab\frac{a-1}{b-1}\cdots\frac{a-(k-1)}{b-(k-1)}\le\Big(\frac ab\Big)^k.
\end{equation*}
So,
\begin{equation*}
r_m\le\Big(\frac{m+3n}{m+4n}\Big)^{m+n}\le\tr_m:=\tr_{m,n}:=\exp-\frac{(m+n)n}{m+4n}.
\end{equation*}

Next, for $m,n$ as before,
\begin{equation*}
\frac{\rho_{m+1}}{\rho_m}-1
=-n\frac pq<0,
\end{equation*}
where
\begin{equation*}
\begin{aligned}
p&:=22 + 270 m + 920 m^2 + 1184 m^3 + 512 m^4 + 155 n + 1155 m n +
2328 m^2 n \\
&+ 1376 m^3 n + 330 n^2 + 1438 m n^2 + 1328 m^2 n^2 +
265 n^3 + 529 m n^3 + 68 n^4,
\end{aligned}
\end{equation*}
\begin{equation*}
q:=(1 + 4 m + 3 n) (2 + 4 m + 3 n) (3 + 4 m + 3 n) (4 + 4 m + 3 n) (1 +
m + 4 n).
\end{equation*}
So, $\rho_m$ is decreasing in $m\ge1$. Also, it easy to see that $\tr_m$ is decreasing in $m\ge1$.

So, for all $m\ge1$ and $n\ge6$
\begin{equation*}
\begin{aligned}
r_m+\rho_m &\le \tr_1+\rho_1 \\
&
\begin{aligned}
=s(n)&:=\exp\Big(-\frac{(1+n)n}{1+4n}\Big) \\
&+\frac{8 (n+1) (2 n+1) (4 n+3)}{3 (3 n+1) (3 n+2) (3 n+4)}
%\le s_6=0.970\ldots
<1,
\end{aligned}
\end{aligned}
\tag{30}\label{30}
\end{equation*}
so that \eqref{20} holds, as desired. The inequality $s(n)<1$ for $n\ge6$ in \eqref{30} holds because for $t(n):=(s(n)-1)\exp\frac{(1+n)n}{1+4n}$ we have $t(6)=-0.160\ldots<0$ and
\begin{equation*}
t'(n)=-\frac PQ\,\exp\frac{(1+n)n}{1+4n}<0,
\end{equation*}
where
\begin{equation*}
P:=176 + 2304 n + 11124 n^2 + 25740 n^3 + 32471 n^4 + 26908 n^5 +
19299 n^6 + 10062 n^7 + 1836 n^8,
\end{equation*}
\begin{equation*}
Q:=3 (1 + 3 n)^2 (2 + 3 n)^2 (4 + 3 n)^2 (1 + 4 n)^2,
\end{equation*}
so that $t(n)<0$ for $n\ge6$.

The cases of $n\in\{1,\dots,5\}$ are substantially easier. For each of these $5$ cases, $r_m+\rho_m$ is a certain rational expression in $m$, and then \eqref{20} fails to hold only for $(m,n)\in\{(1,1),(2,1),(1,2),(1,3)\}$ -- but for such $(m,n)$ the desired inequality in the OP is checked by a simple direct calculation. $\quad\Box$