Czech-Polish-Slovak Mathematical Match competition problems

We prove that $c$ satisfies the condition in the statement if and only if $c \le 2$.

Let us first consider any $c \le 2$. For any positive integer $k$, define
$$n = k^2 + 1 \quad \text{and} \quad m = (k-1)^2 + 1$$
Observe that
$$m + c\sqrt{m-1} + 1 \le k^2 - 2k + 2 + 2(k-1) + 1 = k^2 + 1 = n$$
and
$$(n, n+1, \dots, 2n-m) = (k^2+1, k^2+2, \dots, k^2+2k).$$
Therefore, every such pair $(n, m)$ indeed satisfies the property from the problem statement, and there are infinitely many such pairs.

Now let us consider any $c > 2$, and let $(n, m)$ be any pair of positive integers satisfying the property from the problem statement. Observe that for each positive integer $n$, the number $\lceil\sqrt{n}\rceil^2$ is always between numbers $n$ and $(\sqrt{n}+1)^2$ (inclusive), hence there is always a square of an integer in the range
$$n, n+1, \dots, n + \lfloor 2\sqrt{n} \rfloor + 1.$$
This implies that $2n - m < n + \lfloor 2\sqrt{n} \rfloor + 1$, so in particular
$$m \ge n - 2\sqrt{n}.$$
Combining this with the inequality from the problem statement yields
$$n \ge n - 2\sqrt{n} + c\sqrt{n - 2\sqrt{n} - 1} + 1. \quad (1)$$
Observe that since $c > 2$, we have $c\sqrt{n - 2\sqrt{n} - 1} > 2\sqrt{n}$ for large enough $n$. Indeed, equivalently we have $1 - \frac{2}{\sqrt{n}} - \frac{1}{n} > \frac{4}{c^2}$, and the left-hand side tends to 1 as $n$ grows to infinity while the right hand side is strictly smaller than 1. This implies that (1) may be satisfied only for finitely many positive integers $n$. Since $m \le n$ for all pairs $(n, m)$ satisfying the conditions from the problem statement, this implies that there are only finitely many such pairs $(n, m)$. $\square$