Find all natural numbers that satisfy the condition of the given problem
Solution 1:
The solution you found works formally, but not in practice. It works for $b=2,3$ to give $12,33$, but larger primes ($b>3$) have the form $b=6k\pm 1$ so their squares have the form $b^2=6m+1 \Rightarrow b^2+2=6m+3$.
Hence, for $b>3$, we have $3\mid (b^2+2)$ and $b$ is no longer the smallest proper divisor.