Solution:

We shall prove that $n+1$, $n+2$ and $n+3$ are prime powers. Assume the contrary and let some of them has the form $ab$, where $a \geq 2$, $b \geq 2$ and $(a, b) = 1$. Since $ab$ does not divide $M$, then $a$ or $b$ does not divide $M$. Let $a$ does not divide $M$. Then it follows that $a \geq n+1$, i.e. $ab - a \leq 2$. Since $ab - a = a(b-1) \geq 2$ for $a \geq 2$, $b \geq 2$, it follows that $a = 2$, $b = 2$, which contradicts $(a, b) = 1$.

Therefore $n+1$, $n+2$ and $n+3$ are prime powers. At least one of them is even, so it has the form $2^{x}$. Analogously, at least one of them is divisible by $3$, so it has the form $3^{y}$. By parity arguments we conclude that
$$2^{x} = 3^{y} \pm 1$$

Case 1. Let $2^{x} = 3^{y} + 1$. Since $2^{x} \equiv 1 \pmod{3}$ for $x$ even and $2^{x} \equiv 2 \pmod{3}$ for $x$ odd, we see that $x = 2z$, where $z$ is a nonnegative integer, and $(2^{z} - 1)(2^{z} + 1) = 3^{y}$. Then $2^{z} - 1$ and $2^{z} + 1$ are powers of $3$, which is possible only for $z = 1$. Therefore $3^{y} = 3$, $2^{x} = 4$, whence $n = 1$ or $n = 2$. These solutions are achieved for $M = 1$ and $M = 2$, respectively.

Case 2. Let $2^{x} = 3^{y} - 1$. We may assume that $x \geq 2$ because for $x = 1$ we obtain one of the above answers for $n$. We have $3^{y} \equiv 1 \pmod{4}$ for $y$ even and $3^{y} \equiv 3 \pmod{4}$ for $y$ odd. Therefore $y = 2z$ and $2^{x} = (3^{z} - 1)(3^{z} + 1)$. Then $3^{z} - 1$ and $3^{z} + 1$ are powers of $2$ which is possible only for $z = 1$. Therefore $3^{y} = 9$, $2^{x} = 8$, whence $n = 6$ or $n = 7$. The solution $n = 6$ is achieved, for instance, if $M = 60$. For $n = 7$ we obtain $n+3 = 10$, which is not a prime power.

Finally, the solutions are $n = 1$, $n = 2$ and $n = 6$.