Is this a Valid proof for $(2n+1,3n+1)=1$?
Yes, nice job! That's a very nice application of the method that I described in your prior question. Notice again how simple the problem becomes once it is reduced to equational (congruence) form.
Suppose $d$ is a positive integer dividing both $2n+1$ and $3n+1$.
$d$ also dividing $3(2n+1)$ and $2(3n+1)$.
$d$ dividing $(6n+3)-(6n+2)=1$.
$d$ dividing $1$.
$d=1$