Claim. For any positive integer $a$ and real number $x > 0$ the following inequality holds:
$$a(x+1)^{\frac{1}{a}} \le x+a,$$
where the equality is satisfied if and only if $a=1$.

Proof. If $a=1$, we easily see that the equality holds for all real numbers $x > 0$. Let us assume that $a > 1$, and let us apply the inequality between arithmetic and geometric means for $x+1$ and $a-1$ copies of $1$:
$$\frac{(x+1)+\overbrace{1+1+\cdots+1}^{a-1}}{a} \ge \sqrt[a]{(x+1) \cdot 1^{a-1}},$$
or equivalently,
$$a(x+1)^{\frac{1}{a}} \le x+a.$$
Notice that the equality is attained only if the equation $x+1=1$ is satisfied, which is impossible due to the assumption $x > 0$. $\square$

Let us now prove the original problem, using the above claim. We have
$$\begin{aligned} M(x+1)^k &= a_1(x+1)^{\frac{1}{a_1}} a_2(x+1)^{\frac{1}{a_2}} \cdots a_n(x+1)^{\frac{1}{a_n}} \\ &\le (x+a_1)(x+a_2)\cdots(x+a_n), \end{aligned}$$
where equality holds in case $a_1 = a_2 = \cdots = a_n = 1$. However, $a_1 a_2 \cdots a_n = M > 1$ so we conclude that this is not the case. Therefore, we conclude that the original equation has no solutions in positive real numbers.