What is 2017 doing here?

$\begin{aligned} 44^{\log_{11}121} + 6561^{\log_{49}7} & = 6605^{\log_X 2017} \\ 44^{\log_{11}11^2} + 6561^{\log_{49}\sqrt{49}} & = 6605^{\log_X 2017} \\ 44^2 + \sqrt{6561} & = 6605^{\log_X 2017} \\ 1936 + 81 & = 6605^{\log_X 2017} \\ 2017 & = \color{#3D99F6} 6605 ^{\log_X 2017} & \small \color{#3D99F6} \text{Note that } m^{\log_b n} = n^{\log_b m} \\ 2017 & = \color{#3D99F6} 2017^{\log_X 6605} \\ \implies \log_X 6605 & = 1 \\ X & = \boxed{6605} \end{aligned}$

Therefore, the answer is $\boxed{\text{None of the others}}$ .

$\large { 44 }^{ \log _{ 11 }{ 121 } } + { 6561 }^{ \log _{ 49 }{ 7 } } = \large { 44 }^{ \log _{ 11 }{ 11^{2} } } + { 6561 }^{ \log _{ 49 }{ 49^{0.5} } } = 44^{2} + \sqrt{6561} = 2017.$ If we now have $\large 2017 = { 6605 }^{ \log _{ X }{ 2017 } }$ , then only $X = 6605$ solves this equation.