It is an immediate generalization of a famous Czech Problem from 2000. See here.

For $n=m=1$, we have equality. I will prove that $(1-x_1x_2\cdots x_n)^m+(1-y_1^m)(1-y_2^m)\cdots(1-y_n^m)\ge 1$. Let's prove this by induction over $n$. Check that it is true for $n=1$. For $n=2$, we have $(1-(1-y_1)(1-y_2))^m+(1-y_1^m)(1-y_2^m)\ge 1\Leftrightarrow (y_1+y_2-y_1y_2)^m\ge y_1^m+y_2^m-(y_1y_2)^m\Leftrightarrow (y_1+y_2-y_1y_2)^m+(y_1y_2)^m\ge y_1^m+y_2^m$. Let $y_1+y_2=A$ and suppose that $y_1\ge y_2$. Then, $\frac A2\ge y_2>y_1y_2>0$. Let $f(x)=(A-x)^m+x^m$. We have $f'(x)=mx^{m-1}-m(A-x)^{m-1}=m(x^{m-1}-(A-x)^{m-1})\leq 0$ where $\frac A2\ge x>0$. Hence, $f$ is non-increasing in the interval $\left[0,\frac A2\right]$. Since $\frac A2\ge y_2>y_1y_2>0$, we have $f(y_1y_2)\ge f(y_2)$. Thus, $(y_1+y_2-y_1y_2)^m+(y_1y_2)^m\ge y_1^m+y_2^m$, as desired. Assume that $(1-x_1x_2\cdots x_n)^m+(1-y_1^m)(1-y_2^m)\cdots(1-y_n^m)\ge 1$ where $n\ge 2$. Let's show that $(1-x_1x_2\cdots x_{n+1})^m+(1-y_1^m)(1-y_2^m)\cdots(1-y_{n+1}^m)\ge 1$. $(1-(x_1x_2)\cdots x_{n+1})^m+(1-y_1^m)(1-y_2^m)\cdots(1-y_{n+1}^m)\ge \left[1-(1-(1-x_1x_2)^m)(1-y_3^m)(1-y_4^m)\cdots (1-y_{n+1}^m)\right]+$ $(1-y_1^m)(1-y_2^m)\cdots(1-y_{n+1}^m)\ge \left[1-(1-(1-(1-y_1^m)(1-y_2^m)))(1-y_3^m)(1-y_4^m)\cdots (1-y_{n+1}^m)\right]+(1-y_1^m)(1-y_2^m)\cdots(1-y_{n+1}^m)=$ $\left[1-(1-y_1^m)(1-y_2^m)(1-y_3^m)(1-y_4^m)\cdots (1-y_{n+1}^m)\right]+(1-y_1^m)(1-y_2^m)\cdots(1-y_{n+1}^m)=1$, as desired. Hence completes the proof.