We will assume that by number we mean positive integer.
Call a number of the form $3n+2$ which has no prime divisor of the form $3m+2$ bad. We will show that there are no bad numbers.
Suppose to the contrary that there is at least one bad number. Then there is a smallest bad number $b$.
It is clear that $b\ne 1$. It is also clear that $b$ is not prime. For if $b$ is prime, then it has a prime divisor of the form $3m+2$, namely itself. So there are integers $k$ and $l$, both bigger than $1$, such that $b=kl$.
If one of $k$ or $l$ is divisible by $3$, then so is their product. But $b$ cannot be divisible by $3$. For if $3$ divides $3n+2$, then $3$ must divide $2$, which is not the case.
If both $k$ and $l$ are congruent to $1$ modulo $3$, then so is their product. But $b$ is congruent to $2$ modulo $3$.
Thus at least one of $k$ and $l$ (in fact exactly one) is congruent to $2$ modulo $3$. Let it be $k$.
Because $k\lt b$, by the definition of $b$ as the smallest bad number, $k$ cannot be bad. It follows that $k$ is divisible by a prime $p$ of the form $3m+2$. But then so is $b$, contradicting the fact that $b$ is bad.
Remark: A proof that uses the fact that every integer $\gt 1$ is a product of primes is more efficient. Note that we do not need the Fundamental Theorem of Arithmetic, aka the Unique Factorization Theorem. It is enough to have a factorization.