From the definition, the numbers $n = 1, 4, 9$ satisfy the required property. Suppose that for some $t \in \mathbb{Z}^+$, the number $n = t^2 \geq 15$ also satisfies the required property. We have two cases: Case 1. If $n$ is a power of $2$ Then $n \in {16, 64}$ as $2^4 + 15 = 31, 2^5 + 15 = 47$, and $2^6 + 15 = 79$ are prime, and $2^7 + 15 = 143 = 11 \cdot 13$ is not a prime power and therefore no higher power of $2$ satisfies the required property. Case 2. Suppose $n$ has some odd prime divisor $p$. If $p > 3$ then $p^2 \mid n$ and $p^2 > 15$ which implies 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 $q = 2$. Or $p^2 + 15 = 2^m$. Considering the above modulo $3$, we can see that $p^2 + 15 \equiv 0, 1 \mod 3$, so $2^m \equiv 1 \mod 3$, and so $m$ is even. Suppose $m = 2k$ for some $k \in \mathbb{Z}^+$. So we have $(2^k-p)(2^k + p)=15$ and $(2^k + p)-(2^k-p)= p \geq 10$. Thus $2^k-p = 1$ and $2^k + p = 15$ 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 $y \leq 1$. The numbers $18$ and $21$ are also not nice, so similarly, $x, y$ and $y, z$ cannot both positive. Hence, we just need to consider $n = 4^x \cdot 49^z$ with $z \geq 1$. Note that $73$ 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, $n = 1, 4, 9, 16, 49, 64, 196$.

We exclude all numbers <= 15, because they aren't useful here. If n = k^2 is power of 2, then 16 <= k^2 <= 64 as for p=128, p+15 is composite. So 16, 64 sufficient in this case. If k^2 = 2^a3^b form, then clearly 36 must divide k^2 and 36 isn't nice. Now without loss of generality let property be an odd prime greater than 3 dividing n. As above mentioned, we have that n = 4^a9^b49^c. But 21 isn't nice, so either 9 power or 49 power doesnt exist. Also, 36 isn't nice so either 4 power or 9 power doesn't exist. From this,this could mean k^2 is of form 3^n or 4^a49^b. But 3^n leads to no such numbers since 27 is not nice. Observe that 343 is not nice so b = 1 since powers of two was covered already. We had earlier establish that a = 0, 1, 2, 3. So, from this we get that k^2 = 49, 196 are extra solutions since 407 = 49*8+15 is divisible by 11 and 37. @above I believe we need to exclude 1, 4, 9. a > 15 must divide n.

Lets consider this: Let p be a prime >=15. We take a=p^r - 15. We also take n such that p^r - 15|n. So, by this consideration, there are infinity many n in the form (p^r - 15) ^2*m^2k. Or am I wrong.

@sttsmet why does it work for another divisors of $n$? You considered only $a$, but $m$ has divisors too and perhaps there exists $d|am$ and $d$ doesn’t divide $a$ and $m$. Why does the condition hold for them?

@pixi you are right!! I misunderstood the problem and i thought not all the divisors of n should have this property!