Solution:

Let $a=\log_{2023}7$ and $b=\log_{2023}289.$ Since $2023=7\cdot 289,$ we have $a+b=1.$ Writing $t=\log_{2023}x,$ each logarithm becomes a reciprocal, and the equation turns into $$(1+t)=(a+t)(b+t).$$

Expanding and using $a+b=1,$ the linear terms cancel, leaving $t^2+(ab-1)=0.$ Its two roots satisfy $t_1+t_2=0.$

The corresponding solutions multiply to $x_1x_2=2023^{t_1}\cdot 2023^{t_2}=2023^{\,t_1+t_2} =2023^0=1.$

Thus, the correct answer is C.