Existence of $\{a_{n}\} $ and $\{b_{n}\}$ such that $a_{n}(a_{n}+1)|(b^2_{n}+1)$ 
Show that: there exist two sequences $\{a_{n}\}$ and $\{b_{n}\}$ that are monotonically increasing (or $a_{n+1}>a_{n},b_{n+1}>b_{n},\forall n\in N^{+}$) and for any positive integer $n$ 
  $$a_{n}(a_{n}+1)|(b^2_{n}+1)$$

I assumed $a_{n}=n$, but cannot calculate $b_{n}$ for this case.
How can we prove the existence?
 A: Suppose that $b^2 + 1 = k\cdot a(a+1)$ for $k>1$.  We can rewrite this as follows:
$$(2b)^2 - k(2a+1)^2 = -(k+4)$$
In other words, the pair $(2b,2a+1)$ is the solution to a generalized Pell's equation.  If we can find one solution for some fixed (non-square!) $k$, we can find infinitely many.
We can start with the simplest example, $a=1$, $b=3$, which gives $k=5$, and the solution pair $(6,3)$ with $6^2-5\cdot 3^2 = -9$.
We also want to find the fundamental solution for $x^2-5y^2 = 1$, namely $(9,4)$.
Then we can build a sequence with the desired properties by simply defining $2b_n + (2a_n+1)\sqrt{5} = (6+3\sqrt{5})(9+4\sqrt{5})^n$ for $n\geq 0$.
(To see that these sequences really satisfy the original equation, multiply each side by its conjugate.)
Note that $n=1$ gives vadim123's example.  Neil's examples could be used to construct infinitely many solutions for $k=13$ and $97$, respectively.  It would be interesting to determine for what $k$ a solution exists, but this seems potentially intractable.

Without going into the details, we can also define these two sequences by straightforward linear recurrences:
$$a_0 = 1, a_1 = 25, a_{n+2} = 18 a_{n+1} - a_n + 8$$
$$b_0 = 3, b_1 = 57, b_{n+2} = 18 b_{n+1} - b_n$$

Other observations:
By setting $a=1$, $b=2t+1$, we can see that any $k=2t^2+2t+1$ produces infinitely many solutions.  This accounts for the examples of $k=5$ and $k=13$, and predicts nontrivial examples for $k=41$, but it doesn't explain the $k=19$ example.
(Actually, I just discovered that the solution set of a generalized Pell's equation can have more than one fundamental solution, so the last paragraph actually doesn't explain Neil's $k=13$ example, but rather generates a kind of parallel family of examples.)
For $k=41$, incidentally, the first nontrivial example is $5953 \cdot 5954 \mid (38121^2+1)$, which I produced by multiplying the $a=1$, $b=9$ solution $18+3\sqrt{41}$ by the fundamental solution $5850 + 915\sqrt{41}$.
