Say that $n^2+np$ is a perfect square.
$$ \implies n^2+np = k^2 \implies n(n+p) = k^2 $$
Since $p$ is a prime number, there can be one of the two cases:
(A) $p$ does not divide $n$, in which case gcd$(p,n)=1$. Therefore, gcd$(n, n+p)=1$. Since $n(n+p)$ is a perfect square and $n$ and $n+p$ are co-prime, therefore, $n$ and $n+p$ are perfect squares. Therefore, assume that
$$ n = k^2, \ n+p = m^2 $$
$$ p = (m^2-k^2) = (m-k)(m+k) $$
Since $p$ is prime, it forces $m-k$ to be $1$, that is, $m-k=1$. Therefore, we have $p=m+k = 2k+1.$ And therefore, $n = k^2= \left( \frac{p-1}{2} \right)^2, $ which is unique.
(B) $p$ divides $n$ that is $n = Ip$ for some positive integer $I>1$. It is trivial to see that for $n=p $ the expression is not a perfect square. So, the given equality becomes
$$ Ip (Ip+p) = k^2 \implies p^2 I(I+1) = k^2 $$
For this to be true, $I(I+1)$ must be a perfect square. But for every $I>0$, we have
$$ I^2 < I(I+1)<(I+1)^2 $$
and therefore can never be a perfect square. Hence, this case is ruled out.
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