$$
\sqrt{n+\sqrt{n+\sqrt{n+\ldots \sqrt{n}}}}=m
$$

Let's find the integer solutions of the above equation!
For a given non-negative $n$, we denote by $n_{k}$ the quantity $\sqrt{n+\sqrt{n+\ldots+\sqrt{n}}}$, where the number of square roots is $k (k \geq 1)$.

If $n_{k}$ is not an integer for some $k$, then $n_{k} + n$ is also not an integer, and thus $n_{k+1} = \sqrt{n + n_{k}}$ is not an integer either, since the square of an integer would be an integer. This implies that $n_{j}$ is not an integer for any $j > k$.

We now distinguish three cases:

1. If $n$ is an integer and not a perfect square, then $n_{1}$ is irrational and thus not an integer. In this case, none of the terms in the sequence $n_{1}, n_{2}, \ldots, n_{k}, \ldots$ are integers, so the equation has no integer solution regardless of the number of square roots on the left side.
2. If $n = 0$, then $n_{k} = 0$ for all $k$. The equation has the solution $n = m = 0$ regardless of the number of square roots on the left side. (The so-called trivial solution.)
3. If $n \geq 1$ and is a perfect square, let $n = r^{2}$, where $r \geq 1$ is an integer. Then

\[
r < \sqrt{r^{2} + r} = \sqrt{n + r} = \sqrt{n + \sqrt{n}} = n_{2} < r + 1
\]

so $n_{2}$ cannot be an integer. According to the introductory remark, $n_{3}, n_{4}, \ldots, n_{k}, \ldots$ are also not integers, so if the number of square roots is two or more, there is no integer solution to the equation for $n = r^{2} \geq 1$. In the case of a single square root, we get the equation $\sqrt{n} = m$, which has the solution $(m^{2}, m)$ for every integer $m \geq 1$.

In summary: The only integer solution to the equation is the pair $n = m = 0$. Additionally, there is a solution only if there is a single square root on the left side, in which case every pair $(m^{2}, m)$, where $m \geq 1$ is an integer, is a solution.