False as stated if $M=\{2017\}$.

Sorry, I make mistake in translation. I changed problem.

Assume that $ M$ does not contain any composite numbers. Let $ a$ and $ b$ be two different elements of $ M$ such that $ a >3$ and $ b >2$. Note that $ a$ is a prime number larger than $ 3$ and $ b$ is an odd positive integer. We consider the following two cases (1)(2) : (1) $ \, a\equiv 1\pmod 6$ Since $ a^{b} +2\equiv 0\pmod 3$, we must have $ a^{b} -2\in M$. However, $ \left( a^{b} -2\right)^{a} -2\equiv 0\pmod 3$ and $ \left( a^{b} -2\right)^{a} +2\equiv 0\pmod a$. (2) $ \, a\equiv -1\pmod 6$ Since $ a^{b} -2\equiv 0\pmod 3$, we must have $ a^{b} +2\in M$. However, $ \left( a^{b} +2\right)^{a} +2\equiv 0\pmod 3$ and $ \left( a^{b} +2\right)^{a} -2\equiv 0\pmod a$. Hence, $ M$ must have a composite number. $ \blacksquare$