If I'm not wrong, writing $x_{n+2} - x_{n+1} - x_{n} = 2\sqrt{x_{n}x_{n+1}+121} $, squaring, writing another one for the next index, subtracting and factorizing, yields $(x_{n+3} - x_n)(x_{n+3} - 2x_{n+2} - 2x_{n+1} + x_n) = 0$. Since $(x_n)_{n\geq 2}$ is obviously increasing, that means $x_{n+3} - 2x_{n+2} - 2x_{n+1} + x_n = 0$. Now, $x_3$ turns to be equal to $70$, thus integer, and by simple induction all terms will turn integer, not just $x_{2013}$

My solution. consider the statement $P(n):(\ x_n$ and $x_{n+1} \in \mathbb{N} $ and $x_nx_{n+1}+121$ is a perfect square $)$ Let's prove P is true for all positive integer n by induction. It is obvious in the base case. suppose $P(n)$ is true, let's prove $P(n+1) $ is true . 1) $P(n)\Rightarrow x_{n+1}\in N$ and $x_{n+2}=x_n+x_{n+1}+2\sqrt{x_nx_{n+1}+121} \in N$ 2) $x_{n+1}x_{n+2}+121=$ $x_nx_{n+1}+121+x_{n+1}^2+2x_{n+1}\sqrt{x_{n+1}x_n+121}=$ $(\sqrt{x_nx_{n+1}+121}+x_{n+1})^2 $ which is a perfect square. Hence $P(n)$ is true, which yields $x_n\in $ for all positive integers $n$.