I'm still trying, Minkowski Theorem might work.

OK, let me post my solution: Suppose $\triangle ABC$ is a lattice triangle that contains one $m$-lattice point $P$. I will show that if $S_{\triangle ABC}\ge 12m^3$, then there will be another $m$-lattice point inside $\triangle ABC$. Suppose the barycentric coordinate of $P$ is $(x, y, z) (x\ge y\ge z >0, and x+y+z=1)$. Obviously, if $x\le \frac{m}{m+1}$, then the point$Q$ which satisfies $\vec PQ=m\vec AP$ is another $m$-lattice point in $\triangle ABC$. So now we suppose $x<\frac{m}{m+1}$, which means $y\ge \frac{1}{2(m+1)}$ (this provides space for us to find $m$-lattice points) Let $D,E$ incide $\triangle ABC$ such that $\vec DP=\vec PE=\frac{\vec AB}{2(m+1)}$.Let $F$ lies in $BC$ and $G$ lies in $AC$ such that $\frac{FC}{BC}=\frac{GC}{AC}=\frac{1}{m+1}$.Then $DEFG$ is a parallelogram and $P$ is the midpoint of $DE$. Since $z\le\frac{1}{3}$, we have: $S_{DEFG}=\frac{2}{m+1}(1-z-\frac{1}{m+1})S_{\triangle ABC} \ge \frac{2}{m+1} (\frac{2}{3}-\frac{1}{m+1}) S_{\triangle ABC}\ge \frac{S_{\triangle ABC}}{6m} \ge 2m^2$. Consider another parallelogram $FGHI$ that $P$ is its center.$S_{FGHI}\ge 4m^2$.By Minkowski's Theorem, there is another $m$-lattice inside $FGHI$ which means another $m$-lattice inside $DEFG$.END!

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