So I argue by contradiction as follows.
Assume there exists a face in which there are more than three edges, Then the face must be a cycle, IF the face is four edges, then we can one edge and this will divide it into two triangle, if the face has 5 edges then it's a pentagon and if wee add two edges we will divide it into three triangles, if the face has six edges then we can add three edges and this will divide the face into four triangles. In general, For a face with $n >3$ edges, We can add $n-3$ edges to divide it into triangles.
If there exists a face with less than three edges , then we just have maybe two edges (THIS IS NOT EVEN A FACE ?). Right because the least number of edges required for a face is three because I am not dealing here with multi-edges (and so avoiding loops).
But I feel my argument is not rigorous enough, How can I improve and make it more convincing.
You basically have the idea for the proof.
Let graph $G$ be simple and that $V(G) \geq 3$. Assume we have a non-triangulated face $F$. Since $V(G) \geq 3$, we have $|V(F)| \geq 3$. Then since $F$ is non-triangulated, there exists three vertices $x,y,z \in V(F)$ where $xy \in E(F)$, $yz \in E(F)$ and $xz \not \in E(G)$. Then we can add in the edge $xz$ and still have a planar graph. So for every non-triangluated face we can increase the number of edges and preserve planarity, thus the maximum number of edges is obtained when every face is triangulated.