, 2022

If a positive integer $d$ divides both $m-k^2$ and $n+2k$, $d$ also divides $4(m-k^2)-(n-2k)(n+2k) = 4m-n^2$. Thus when $4m \neq n^2$ holds, $N = |4m-n^2|$ satisfies $d \le N$ for any positive integer $d$ and $(m,n)$ does not satisfy the condition.
On the other hand, when $4m = n^2$ holds, $n = 2t$ for a positive integer $t$ since $n$ is even. Then we have $m = \frac{n^2}{4} = t^2$. Let $k = N$ and $d = t + N$, then $d > N$ holds and
$$\frac{m-k^2}{d} = \frac{t^2-N^2}{t+N} = t-N, \quad \frac{n+2k}{d} = \frac{2t+2N}{t+N} = 2$$
are both integers. Hence $(m,n)$ satisfies the condition.
From the above, it follows that pairs $(m,n)$ satisfying the condition are the pairs which can be written as $(t^2, 2t)$ with a positive integer $t$ and $1 \le t^2, 2t \le 2022$. $1 \le t^2 \le 2022$ shows $1 \le t \le 44$, and then both $t^2$ and $2t$ are integers between 1 and 2022. Hence the answer is 44.