$n + 1$

We will prove that $k \leq n + 1$ is always possible We will use induction on $n$. Base case $n = 2$ is obvious. Now assume it is possible for $n = m$ we will prove it is possible for $n = m + 1$ as well - If there is an empty row switch that row with the bottom row, then switch a column with at least 1 marked square with the rightmost column. Consider the top left $m \times m$ table. Clearly, the number of marked squares there is $\leq m + 1$. Since the bottom row is empty and the rightmost column won't really matter, by induction hypothesis possible to arrange the $m \times m$ table so that it satisfies the condition, and automatically makes the $m + 1 \times m + 1$ table satisfy as well - If there are no empty rows There will be a row with exactly 1 marked square. Switch this row with the bottom row, then switch the column containing this marked square with the rightmost column. Similarly, by induction hypothesis, it is indeed possible.

for $k = n + 2$, mark the squares in the main diagonal and the other two corner squares. To prove this is impossible, note that no matter the arrangement there will always be 2 rows and columns with 2 marked squares, and n - 2 rows and columns with 1 marked square. Assume that it is possible and it will be easy to find a contradiction.