What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$

Note that $\gcd(n, n+1) = \gcd(n, 1) = 1$ do eithet $n=p^2, n+1 = 1$ or $n+1 = p^2, n = 1$; both of which are impossible.

You are given

$$n(n+1) = p^2 \tag{1}\label{eq1}$$

with $n \in \mathbb{N}$ (I'm assuming positive integers is meant here). Note that $n^2 \lt n^2 + n = n(n+1) \lt n^2 + 2n + 1 = (n+1)^2$. Thus, $n \lt p \lt n + 1$, so there are no $p \in \mathbb{N}$ that solve \eqref{eq1}.