Show that infinitely many positive integer pairs $(m,n)$ exist s.t $\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{N}$

I shall find all pairs $(m,n)$ of positive integers such that $$\frac{m+1}{n}+\frac{n+1}{m}\in\mathbb{N}.$$ Let $k$ be such a positive integer for which $$\dfrac{m+1}{n}+\dfrac{n+1}{m}=k\tag{1}$$ for some $m,n\in\mathbb{N}$. Note that $t=m$ is a solution to $$t^2-(kn-1)t+(n^2+n)=0.$$ However, there is another root $t=kn-1-m=\dfrac{n^2+n}{m}$, which is an integer as $kn-1-m\in\mathbb{Z}$, and which is positive since $\dfrac{n^2+n}{m}>0$. Thus, if $(m,n)$ is a positive integer solution to (1), then $(n,kn-1-m)=\left(n,\dfrac{n^2+n}{m}\right)$ is also a positive integer solution.

Now, suppose that $(m_0,n_0)$ is a solution to (1) such that $m_0\geq n_0$ and $m_0+n_0$ is smallest possible. If $m_0>n_0$, we see that $\left(n_0,\dfrac{n_0^2+n_0}{m_0}\right)$ is also a solution, but $$n_0+\frac{n_0^2+n_0}{m_0}=n_0+n_0\left(\frac{n_0+1}{m_0}\right)\leq n_0+n_0<m_0+n_0\,.$$ This contradicts the minimality of $m_0+n_0$, and so $m_0=n_0$ must hold. Thus, $$k=\frac{m_0+1}{m_0}+\frac{m_0+1}{m_0}=2+\frac{2}{m_0}\,.$$ That is, $(m_0,n_0)=(1,1)$ (which gives $k=4$), or $(m_0,n_0)=(2,2)$ (which gives $k=3$).

In the first case, $m_0=n_0=1$ and $k=4$. Define a sequence $(a_0,a_1,a_2,\ldots)$ by taking $a_0=1$, $a_1=1$, and $$a_{r}=4a_{r-1}-a_{r-2}-1$$ for $r=2,3,4,\ldots$. It follows that all solutions $(m,n)$ with $k=4$ such that $m\leq n$ are of the form $(a_{r},a_{r+1})$ for some $r=0,1,2,\ldots$. For example, $a_2=2$, $a_3=6$, $a_4=21$, and $a_5=77$.

In the second case, $m_0=n_0=2$ and $k=3$. Define a sequence $(b_0,b_1,b_2,\ldots)$ by taking $b_0=2$, $b_1=2$, and $$b_r=3b_{r-1}-b_{r-2}-1$$ for $r=2,3,4,\ldots$. It follows that all solutions $(m,n)$ with $k=3$ such that $m\leq n$ are of the form $(b_{r},b_{r+1})$ for some $r=0,1,2,\ldots$. For example, $b_2=3$, $b_3=6$, $b_4=14$, and $b_5=35$.