$2n+1$ and $n^2+1$ are always coprime or their gcd is $5$ Using a spreadsheet, it can be inferred that when $n≡2[5]$, then $\gcd(n^2+1,2n+1)=5$, else $\gcd(n^2+1,2n+1)=1$.
Indeed, when $n≡2[5]$, $n^2+1$ and $2n+1$ can easily be shown to be multiples of $5$, so their gcd is at least $5$. But then, I can't see how to complete the proof.

As $(a,b)=(a,a-nb),$
$(2(n^2+1), 2n+1)=(2(n^2+1)-n(2n+1), 2n+1)=(2-n,2n+1)=(2-n,2n+1+2(2n-1))=(2-n,5)$
So, $(2(n^2+1), 2n+1)=5$ if $5\mid(2-n)$ i.e., if $n\equiv2\pmod 5$
$\implies (n^2+1, 2n+1)=5$ as $(2,2n+1)=1$ is as $(2n+1)$ odd.
If $n\not\equiv2\pmod 5,(2(n^2+1), 2n+1)=1\implies (n^2+1, 2n+1)=1$

Alternatively, 
We know, $(a,b)\mid (ax+by)$  where $a,b,x,y$ are integers.
So, $(n^2+1, 2n+1)\mid \{n(2n+1)-2(n^2+1)\}\implies (n^2+1, 2n+1)\mid(n-2)$
Again, $(2n+1,n-2)\mid\{2n+1-2(n-2)\} \implies (2n+1,n-2)\mid 5$
If $n\equiv2\pmod 5,2n+1\equiv0 \pmod 5$ and $n^2+1\equiv2^2+1\equiv 0 \pmod 5$
hence, $(n^2+1, 2n+1)=5$
Else, $5\not\mid(n-2)\implies (2n+1,n-2)=1$ and $5\not\mid(2n+1)\implies (n^2+1, 2n+1)=1$
