If $M$ is prime power, let $M = p^{\alpha}$, then $N = (p + 1)^{\alpha} \implies p \mid p + 1$, clear contradiction. So, $M$ is not prime power. Let $M + p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, where $p_1 < p_2 < ... < p_k$. Then, $N = (p_1 + 1)^{\alpha_1} \cdots (p_k + 1)^{\alpha_k}$, we have $p_k \mid (p_1 + 1)(p_2 + 1) \cdots (p_k + 1)$. Say $p_k \mid p_i + 1$, where $1 \leq i \leq k - 1$. $p_{k - 1} + 1 \leq p_k \leq p_i + 1$, but $p_i + 1 \leq p_{k - 1} + 1 \implies p_k = p_{k - 1} + 1 \implies p_{k - 1} = 2, p_k = 3$, and there are no primes below $2$. So, $M = 2^{\alpha} \cdot 3^{\beta}$, and $N = 3^{\alpha}2^{2\beta}$, so $\alpha \leq 2\beta$, $\beta \leq \alpha$. The product of the new multipliers is $3^{2\beta}2^{2\alpha}$, which is clearly divisible by $N$, since $2\beta \geq \alpha$ and $2\alpha \geq 2\beta$. Hence proved.