Let $p$ be a prime such that $p|n^2+3 $and $p|n^4+6 \rightarrow p|n^4+6-n^2(n^2+3)+3(n^2+3)=15$. So $p\in {3,5}.$ If $p=3$ then either $n^2+3$ is a power of $3$ or $n^4+6$ is a power of$3$ (since one of them is odd). In the first case : since $v_3(n^2+3)=1$ $\rightarrow n=0$ a contradiction. In the same way discuss case (2). If $p=5$ so either $n^2+3$ or $n^4+6$ has one of the forms $(5^k,3.5^k)$. If $n^2+3=3^i5^k$$(i=0,1)$ $\rightarrow n^2=2 mod5$ a contradiction. In the same way discuss the second case. So there is no such $n$.

n=3 is the only answer

Let $p$ be the largest prime divisor of $n^2+3$ and also the smallest prime divisor of $n^4+6$. Since $n^4+6=(n^2+3)(n^2-3)+15$ we get that $p | 15$. Let $p=3$. Since $p$ is the smallest prime divisor of $n^4+6$ then $2$ does not divide $n^4+6$. Hence $n \equiv 1$ (mod $2$). Furthermore, since $p$ is the largest prime divisor of $n^2+3$ we have $n^2+3=2^a3^b$ for some natural numbers $a$ and $b$. Consider $n^2+3$ mod $8$ to find that $n^2+3 \equiv 4$ (mod $8$) (remember that $n \equiv 1$ (mod $2$)) which means $a=2$. Since $p|n^2+3$ then $3|n$. Take $n^2+3$ mod $9$ to find that $n^2+3 \equiv 3$ (mod $9$). Hence $b=1$ so $n^2+3=12 \to n=3$. Now let $p=5$ but note that $n^2+3$ is never divisible by $5$ so there are no solutions for this case. Hence the only solution is $n=3$.