Let $a$ be a positive integer such that $np + 1 = a^2$. Note that $a > 1$.
$$\text{We have } a^2 - 1 = np, \text{ i.e. } (a-1)(a+1) = np.$$
Since $p$ is prime, it follows that $p \mid a-1$ or $p \mid a+1$. We treat these two cases separately:

i) Let $p \mid a-1$, i.e. let $a = kp + 1$ for some integer $k$.
$$\text{Then } 1 + np = (kp + 1)^2 = k^2p^2 + 2kp + 1, \text{ i.e. } n = k^2p + 2k.$$
Now we have:
$$n + 1 = k^2p + 2k + 1 = k^2 + 2k + 1 + (p-1)k^2 = (k+1)^2 + (p-1)k^2.$$
Notice that $a > 1$ implies $k > 0$, so $k+1$ and $k$ are both positive integers.

ii) Let $p \mid a+1$, so that $a = kp - 1$ for some integer $k$.
Analogously to the first case, we get $n + 1 = (k-1)^2 + (p-1)k^2$.
It remains to show that $k-1$ is indeed a positive integer, i.e. that we have $k \ge 2$.
If we assume the contrary ($k = 1$), it follows that $n + 1 = p - 1$, i.e. $n = p - 2$.
This contradicts the given condition that $n \ge p - 1$.
We conclude that the assertion holds for any choice of $n$ and $p$ satisfying the conditions of the problem.