If $p$ satisfies the given conditions, then we must have that $2, 3$ simultaneously either be quadratic residues, or nonquadratic residues. From the law of quadratic reciprocity, we have this occurs iff $p \equiv \pm 1, \pm 5 \mod 24$. We can verify that $p=5$ does not satisfy the given conditions, so clearly the minimal $p$ is at least $19$. To verify that $p=19$ satisfies the given equation, note that if $2^n+3^n \equiv 0 \mod 19$, then $(2 \cdot 3^{-1} )^n \equiv -1 \mod 19 \implies (2 \cdot 13)^n \equiv -1 \mod 19 \implies 7^n+1 \equiv 0 \mod 19$, and it is easy to verify that this has no solutions in $n$.

MinatoF wrote: Determine the minimum value of $p> 3$ for which there is no natural number $n> 0$ such that $2^n+3^n\equiv 0\pmod{p} $. Here $p$ is prime? If not we can take $p=4$.

Sorry , $p$ must be prime . Thank you yunxia , post edit