What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$ The following is my approach:

$ n^2+n =p^2$
$   n^2+n+\frac{1}{4} = p^2 + \frac{1}{4}$
$   (n+\frac{1}{2})^2 = p^2+\frac{1}{4}$
$    (n+\frac{1}{2}-p) (n+\frac{1}{2}+p) = \frac{1}{4}$

I am not able to proceed further from here. Any suggestion on what to do next?

You are actually very close to the answer. All you really need to do is multiply both sides of $(n+{1\over2}-p)(n+{1\over2}-p)={1\over4}$ by $4$ and then note that the only integer solutions of $ab=1$ is $a=b=\pm1$. Here is a complete presentation:
Assuming only $n,p\in\mathbb{Z}$, we have
$$\begin{align}
n(n+1)=p^2&\iff4n^2+4n=4p^2\\
&\iff(2n+1)^2=(2p)^2+1\\
&\iff(2n+1)^2-(2p)^2=1\\
&\iff(2n+1-2p)(2n+1+2p)=1\\
&\iff2n+1-2p=2n+1+2p=\pm1\\
&\iff p=0\land n\in\{0,-1\}
\end{align}$$
The requirement $n\in\mathbb{N}$ eliminates $n=-1$, leaving $(n,p)=(0,0)$ as the only solution.
