We prove that $m = 2$. First we show that $m \ge 2$. It suffices for any $s \in (0,1)$ to find five equilateral triangles with combined area greater than $2s$, which can not cover an equilateral triangle $\triangle ABC = \Delta$ of area $1$. Let $A_1B_1C_1$ be an equilateral triangle of area $(1+s)/2$ and vertices on the corresponding sides of $\Delta$. Without loss of generality suppose $2BA_1 \le BC$. Then there exist three equilateral triangles that can not cover any of the segments $BA_1$, $CB_1$ and $AC_1$. Then these triangles and two equilateral triangles $\Delta_1$ and $\Delta_2$ of areas $s$ can not cover $\Delta$. Otherwise $\Delta_1$ and $\Delta_2$ cover points from the given three segments and therefore one of them, say $\Delta_1$, covers points from two of them, say $D \in A_1B$ and $E \in B_1C$. Since $S_{A_1B_1C_1} \ge \frac{1}{3}$, we have that $\not\le A_1B_1C \ge 90^\circ$ (prove!) implying that the side of $\Delta_1$ is at least $DE \ge A_1B_1$. Therefore $S_{\Delta_1} \ge S_{A_1B_1C_1}$, a contradiction.

We prove now that given five equilateral triangles of areas $a^2, b^2, c^2, d^2$ and $e^2$, such that $a^2 + b^2 + c^2 + d^2 + e^2 = 2$, there exist four of them that cover $\Delta$. Let $a \ge b \ge c \ge d \ge e > 0$. If $a \ge 1$ then triangle of area $a^2$ covers $\Delta$. In the opposite case $b+c>1$. This is obvious when $c > 1/2$ (because $b \ge c$), and otherwise
$$b^2 = 2 - a^2 - c^2 - d^2 - e^2 > 1 - 3c^2 \ge (1-c)^2.$$
Therefore the triangles with areas $a^2, b^2$ and $c^2$, cut from the vertices of $\Delta$ intersect each other. They do not cover $\Delta$ if $f = 2 - a - b - c > 0$ and an equilateral triangle of area $f^2$ is not covered. We have to prove that $d \ge f$. This is obvious when $d > 1/2$ (because $a, b, c \ge d$), otherwise it follows from $a, b, c < 1$ that
$$d \ge 2d^2 \ge d^2 + e^2 = 2 - a^2 - b^2 - c^2 > 2 - a - b - c = f.$$