To infect an $m \times n$ board, you need at least
infected cells to start with.
Note: This proof is incorrect, and I don't think it can be easily fixed. Nevertheless I'm leaving this answer up because it shows the difficulty of this straightforward approach.
The proof is by induction.
Base case: It is obviously true that you need $1$ infected square to infect a $1 \times 1$ board. That may seem too trivial to work as a base case (though it is perfectly valid), so if you prefer you can use the fact that a $2\times2$ board needs at least $3$ infected squares to also infect the fourth.
Induction step: Suppose for the sake of induction the hypothesis holds for a particular board size $m \times n$. Add a new row or column to the side of that board. Every newly added square has only a single neighbour that is part of the old board, so if none of the added squares are infected, they can never have two infected neighbours and so will remain healthy. Therefore, to infect the newly added row/column, at least one of the new squares must start off sick. Furthermore, you cannot infect any of the new squares before its adjacent cell on the original board is infected. This means that the original board must be infected without any help from the squares in the new row/column (*). The original board needed at least $m+n-1$ infected squares, so the expanded board must therefore need at least one more, i.e. $(m+n-1)+1$ infected squares.
You can rewrite this as $(m+1)+n-1$ or $m+(n+1)-1$ depending on whether you added a row or a column to the board.
By induction the result follows: $m+n-1$ infected squares are needed on an $m \times n$ board.
The flaw at (*) is:
If you place two or more adjacent infected squares in the extra row/column, then the extra row/column helps infect the rest of the board, and maybe that allows the rest of the board to have fewer initially infected squares to compensate. The argument I used does not disprove the possibility of having an extra infected square in the extra column which then saves 2 or more initial infected squares from the rest of the board.