Let $r=\frac{a}{b},b>0$ is rational and $\sqrt{r+1}$ is rational then $\sqrt{r+1}=\frac{\sqrt{(a+b)b}}{b} \to (a+b)b$ is square $\to b$ is square because $(a+b,b)=1$ If we set $r_0=r=\frac{a}{b},r_i=\sqrt{r_{i-1}+1}$ then to have $r_n$ as rational we need that $b=k^{2^{n-1}}$ for some $k$ and obvious that it is possible only for $b=1$ But then $r=a,r_1=\sqrt{a+1}>1$ and if $r_i \geq 2$ then $r_{i+1}=\sqrt{r_i+1} \leq r_i-\frac{1}{3}$ so for some $k$ we will have $1<r_k<2 \to r_k$ is not integer