First, we may prove $$x_n^2 \geq 2n-1,(n=1,2,\cdots).\tag1$$ Obviously, $(1)$ holds for $n=1$. Assume that $(1)$ holds for $n=k$, then $$x_{k+1}^2=x_k^2+2+\frac{1}{x_k^2}\geq x_k^2+2\geq 2n-1+2=2(n+1)-1,$$ which implies $(1)$ also holds for $n=k+1.$ Thus, by induction, $(1)$ holds for all $n=1,2,\cdots.$
Therefore, $\forall M>0$, $\exists N=[\dfrac{M^2+1}{2}]+1$ such that $$x_n \geq \sqrt{2n-1}>M$$ holds for $n>N$. We are done.
You can get a much simpler proof.
First, you can easily prove by induction that $x_n > 0$ for all $n$. The you deduce that $x_{n+1}-x_n = \frac{1}{x_n} > 0$, so $(x_n)$ is strictly increasing.
If $(x_n)$ would converge, then its limit $l$ would verify $l = l + \frac{1}{l}$. This is impossible. So $(x_n)$ diverges, so its limit is $+\infty$.
Your strategy is valid. However, you needn't have spotted a function $f$ for which $f(x_{n+1})-f(x_n)$ has a positive lower bound. Note that the sequence is strictly increasing, so to diverge it just needs to not have a finite limit $L$. And if such an $L$ existed we'd have $L=L+L^{-1}$, which no finite $L$ satisfies. In other words, $x_{n+1}=x_n+g(x_n)$ with $x\ge x_1\implies g(x)>0$ will always diverge to $\infty$.
Given some $f(x)$ with $f > 0$ and $f' < 0,$ any sequence $$ x_{n+1} = x_n + f(x_n) $$ diverges to $+ \infty$
This can be done in a very brief formal manner, as one of the answers. Or, we can (assume $x_1=1$) say that as long as $x_j < 2,$ we have $x_{j+1} > x_j + f(2),$ so the number of steps until $x_j \geq 2$ has a bound in terms of $\frac{1}{f(2)}.$ Same for $x_j \geq 2$ and reaching $3$
In the first place, we will show that $\{x_n\}$ is an increasing sequence, namely, $x_{n+1}>x_n$ holds for $n=1,2,\cdots$. Consider applying mathematical induction. Since $x_1=1$ and $x_2=2$, then $x_2>x_1$. Assume that $x_{k+1}=x_k+\dfrac{1}{x_k}>x_k,$ which implies $x_k> 0$. Then $x_{k+2}=x_{k+1}+\frac{1}{x_{k+1}}=x_{k+1}+\dfrac{1}{x_k+\frac{1}{x_k}}>x_{k+1}$. By induction, $x_{n+1}>x_n$ holds for all $n$. Now, notice that, any increasing sequence $x_n$ has its limit, which is either a definite number, say $L$, or $+\infty$. If $\lim\limits_{n \to \infty}x_n=L<+\infty$, taking the limits of the recursion，we have $L=L+\dfrac{1}{L}$, which has no real root. Therefore,$\lim\limits_{n \to \infty}x_n=+\infty$. We are done.
