Suppose that $ (3n^2+5n+7,n^2+1)=(5n+4,n^2+1)\ne1\tag{1} $ then either $ (5n+4,n+i)=(4-5i,n+i)\ne1\tag{2} $ or $ (5n+4,n-i)=(4+5i,n-i)\ne1\tag{3} $ Since $4-5i$ is a Gaussian prime, $(2)\Rightarrow4-5i\,|\,n+i$. That is, $ \frac{n+i}{4-5i}=\frac{(4n-5)+(5n+4)i}{41}\in\mathbb{Z}[i]\tag{4} $ which is true if and only if $n\equiv32\pmod{41}$.
Since $4+5i$ is a Gaussian prime, $(3)\Rightarrow4+5i\,|\,n-i$. That is, $ \frac{n-i}{4+5i}=\frac{(4n-5)-(5n+4)i}{41}\in\mathbb{Z}[i]\tag{5} $ which is true if and only if $n\equiv32\pmod{41}$.
Thus, $(1)$ implies either $ (2)\Rightarrow4-5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{6} $ or $ (3)\Rightarrow4+5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{7} $ Therefore, $ (1)\Rightarrow n\equiv32\pmod{41}\tag{8} $ It is easy to verify that $ n\equiv32\pmod{41}\Rightarrow41\,|\,(3n^2+5n+7,n^2+1)\tag{9} $
Again, $(1)$ implies either $ (2)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\tag{10} $ or $ (3)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\tag{11} $ Therefore, $ (1)\Rightarrow41=(4-5i)(4+5i)\,|\,(3n^2+5n+7,n^2+1)\tag{12} $ Finally, as Pambos points out, the Euclidean Algorithm yields $ (15n+13)(n^2+1)-(5n-4)(3n^2+5n+7)=41\tag{13} $ Therefore, $ (3n^2+5n+7,n^2+1)\,|\,41\tag{14} $