First, let us assume that $r>2$, and take a positive integer $a \geq 1 /(r-2)$.
Then, if we let $a_{n}=a+\lfloor n / 2\rfloor$ for $n=1,2, \ldots$, the sequence $a_{n}$ satisfies the inequalities
$$\sqrt{a_{n}^{2}+r a_{n+1}} \geq \sqrt{a_{n}^{2}+r a_{n}} \geq \sqrt{a_{n}^{2}+\left(2+\frac{1}{a}\right) a_{n}} \geq a_{n}+1=a_{n+2}$$
but since $a_{n+2}>a_{n}$ for any $n$, we see that $r$ does not satisfy the condition given in the problem.

Now we show that $r=2$ does satisfy the condition of the problem. Suppose $a_{1}, a_{2}, \ldots$ is a sequence of positive integers satisfying the inequalities given in the problem, and there exists a positive integer $m$ for which $a_{m+2}>a_{m}$ is satisfied.
By induction we prove the following assertion:
(†) $\quad a_{m+2k} \leq a_{m+2k-1}=a_{m+1}$ holds for every positive integer $k$.
The truth of $(\dagger)$ for $k=1$ follows from the inequalities below
$$2 a_{m+2}-1=a_{m+2}^{2}-\left(a_{m+2}-1\right)^{2} \leq a_{m}^{2}+2 a_{m+1}-\left(a_{m+2}-1\right)^{2} \leq 2 a_{m+1}$$
Let us assume that $(\dagger)$ holds for some positive integer $k$. From
$$a_{m+1}^{2} \leq a_{m+2k+1}^{2} \leq a_{m+2k-1}^{2}+2 a_{m+2k} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}$$
it follows that $a_{m+2k+1}=a_{m+1}$ must hold. Furthermore, since $a_{m+2k} \leq a_{m+1}$, we have
$$a_{m+2k+2}^{2} \leq a_{m+2k}^{2}+2 a_{m+2k+1} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}$$
from which it follows that $a_{m+2k+2} \leq a_{m+1}$, which proves the assertion $(\dagger)$.

We can conclude that for the value of $m$ with which we started our argument above, $a_{m+2k+1}=a_{m+1}$ holds for every positive integer $k$. Therefore, in order to finish the proof, it is enough to show that $a_{m+2k}$ becomes constant after some value of $k$. Since every $a_{m+2k}$ is a positive integer less than or equal to $a_{m+1}$, there exists $k=K$ for which $a_{m+2K}$ takes the maximum value. By the monotonicity of $a_{m+2k}$, it then follows that $a_{m+2k}=a_{m+2K}$ for all $k \geq K$.

We only give an alternative proof of the assertion $(\dagger)$ in solution 1. Let $\{a_{n}\}$ be a sequence satisfying the inequalities given in the problem. We will use the following key observations:

a. If $a_{n+1} \leq a_{n}$ for some $n \geq 1$, then
$$a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n}^{2}+2 a_{n}+1}=a_{n}+1$$
hence $a_{n}=a_{n+2}$.

b. If $a_{n} \leq a_{n+1}$ for some $n \geq 1$, then
$$a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n+1}^{2}+2 a_{n+1}+1}=a_{n+1}+1$$
hence $a_{n} \leq a_{n+2} \leq a_{n+1}$.

Now let $m$ be a positive integer such that $a_{m+2}>a_{m}$. By the observations above, we must have $a_{m}<a_{m+2} \leq a_{m+1}$. Thus the assertion $(\dagger)$ is true for $k=1$. Assume that the assertion holds for some positive integer $k$. Using observation (a), we get $a_{m+2k+1}=a_{m+2k-1}=a_{m+1}$. Thus $a_{m+2k} \leq a_{m+2k+1}$, and then using observation (b), we get $a_{m+2k+2} \leq a_{m+2k+1}=a_{m+1}$, which proves the assertion $(\dagger)$.