First notice that $n=0, 1$ does not work, so now lets assume $n \ge 2,$ we can factor the expression as$$7^2(7^{n-2}+3),$$since $7^2$ is a perfect square, we must also have $7^{n-2}+3$ to be a perfect square. If $n\ge 2,$ we will get that is $3\pmod{7},$ but $3$ is not a perfect square $\pmod{7}$