Let $\sqrt{n^2 + pn} = m$, $m \in \mathbb{N}_0$. Then $n^2 + pn - m^2 = 0$.
The solutions of this quadratic equation are
$$n = \frac{1}{2} \left( -p \pm \sqrt{p^2 + 4m^2} \right).$$
The solutions are integers, so $p^2 + 4m^2 = w^2$, for some $w \in \mathbb{N}_0$.
Since $w + 2m \ge 0$ and $w + 2m \ge w - 2m$, from $p^2 = (w - 2m)(w + 2m)$, we obtain the following possibilities:
$$\begin{cases} w - 2m = 1 \\ w + 2m = p^2 \end{cases} \quad \text{or} \quad \begin{cases} w - 2m = p \\ w + 2m = p \end{cases}$$
In the first case we obtain $w = \frac{p^2 + 1}{2}$, $m = \frac{p^2 - 1}{4}$, so it follows:
$$\begin{aligned} n &= \frac{1}{2} \left( -p \pm \sqrt{p^2 + 4 \left( \frac{p^2 - 1}{4} \right)^2} \right) \\ n_1 &= \left( \frac{p-1}{2} \right)^2, \quad n_2 = -\left( \frac{p+1}{2} \right)^2. \end{aligned}$$
These numbers are integers only when $p$ is odd, i.e. only if $p > 2$.

In the second case we obtain $w = p$, $m = 0$, so it follows that $n = \frac{1}{2}(-p \pm p)$ or $n_3 = 0$ and $n_4 = -p$.