Solution:
We call a positive integer $a$ "nice" if $a+15$ is a prime power.
From the definition, the numbers $n=1, 4, 9$ satisfy the required property. Suppose that for some $t \in \mathbb{Z}^{+}$, the number $n=t^{2} \geqslant 15$ also satisfies the required property. We have two cases:

1. If $n$ is a power of $2$, then $n \in\{16, 64\}$ since
$$2^{4}+15=31, \quad 2^{5}+15=47, \quad \text{and} \quad 2^{6}+15=79$$
are prime, and $2^{7}+15=143=11 \cdot 13$ is not a prime power. (Thus $2^{7}$ does not divide $n$ and therefore no higher power of $2$ satisfies the required property.)

2. Suppose $n$ has some odd prime divisor $p$. If $p>3$ then $p^{2} \mid n$ and $p^{2}>15$ which imply that $p^{2}$ must be a nice number. Hence
$$p^{2}+15=q^{m}$$
for some prime $q$ and some $m \in \mathbb{Z}^{+}$. Since $p$ is odd, then $p^{2}+15$ is even, thus we can conclude that $q=2$. I.e.
$$p^{2}+15=2^{m}$$
Considering the above modulo $3$, we can see that $p^{2}+15 \equiv 0,1 \pmod{3}$, so $2^{m} \equiv 1 \pmod{3}$, and so $m$ is even. Suppose $m=2k$ for some $k \in \mathbb{Z}^{+}$. So we have $\left(2^{k}-p\right)\left(2^{k}+p\right)=15$ and $\left(2^{k}+p\right)-\left(2^{k}-p\right)=2p \geqslant 10$. Thus
$$2^{k}-p=1 \quad \text{and} \quad 2^{k}+p=15$$
giving $p=7$ and $k=3$. Thus we can write $n=4^{x} \cdot 9^{y} \cdot 49^{z}$ for some non-negative integers $x, y, z$.

Note that $27$ is not nice, so $27 \nmid n$ and therefore $y \leqslant 1$. The numbers $18$ and $21$ are also not nice, so similarly, $x, y$ and $y, z$ cannot both be positive. Hence, we just need to consider $n=4^{x} \cdot 49^{z}$ with $z \geqslant 1$.

Note that $7^{3}$ is not nice, so $z=1$. By checking directly, we can see that $7^{2}+15=2^{6}$, $2 \cdot 7^{2}+15=113$, $4 \cdot 7^{2}+15=211$ are nice, but $8 \cdot 7^{2}$ is not nice, so only $n=49, 196$ satisfy the required property.

Therefore, the numbers $n$ which satisfy the required property are $1, 4, 9, 16, 49, 64$ and $196$.