Problem 1303

Let us put $n = a d$. The condition $d n + 1 \mid d^2 + n^2$ can be written as $a d^2 + 1 \mid d^2 + a^2 d^2$.
Then $a d^2 + 1$ divides $d^2 + a^2 d^2 - a \cdot (a d^2 + 1) = d^2 - a$ as well.
Let us consider all possible signs of the number $d^2 - a$.
If $d^2 - a > 0$, then it must be $d^2 - a \ge a d^2 + 1$ (multiple is greater than or equal to divisor). Since $a$ is positive, we have $d^2 \ge a d^2 + a + 1 > d^2$, which is impossible.

If $d^2 - a < 0$, then $a - d^2 > 0$. Then it must be $a - d^2 \ge a d^2 + 1$. Since $d$ is positive, we have $a \ge a d^2 + 1 + d^2 > a$, which is also impossible.
Hence, the only possibility is $d^2 - a = 0$, which gives $a = d^2$ and $n = d^3$. In that case $d^4 + 1$ divides $d^2 + d^6 = d^2(d^4 + 1)$, so all possible numbers $n$ are cubes of positive integers.