Denote: $F$ - the number of the faces; $V$ - the number of the vertices; $E$ - the number of the edges.
We use the Euler's polyhedron formula: $F+V=E+2$.
$E=\dfrac{4V}{2}=2V\Longrightarrow V=\dfrac{E}{2}$.
$F+V=E+2\Longrightarrow F=\dfrac{E}{2}+2\quad(1)$.
Denote $f_1, f_2,\dots, f_F$ the faces and let be $e_i$ the number of edges of the face $f_i, 1\le i\le F$.
Each edge belongs to exactly $2$ faces.
Results: $e_1+e_2+\dots+e_F=2E\quad(2)$.
For the triangular faces $f_k$ we have $e_k=3$ and for the other faces $f_j$ we have $e_j\ge 4$.
Using this result in the relation $2$, we obtain:
$2E=e_1+e_2+\dots+e_F\ge 3m+4(F-m)$.
Replacing in the relation $(1)$ results:
$F\ge\dfrac{3m}{4}+F-m+2\Longrightarrow m\ge8$.