The answer is $\boxed{n=0}$ only, which clearly works. Assume for the sake of contradiction that some $n\neq 0$ worked. Then, clearly we must have $n > 0$. Then, $$\frac{5^n-1}{3} \equiv \frac{-1}{3}\equiv 3\pmod{5}.$$This is not a QR mod $5$, so therefore $\frac{5^n-1}{3}$ cannot be a perfect square of an integer. Now, note that since $\frac{5^n-1}{3}$ is an integer, $3\mid 5^n-1$, so $2\mid n$. Thus, letting $n = 2m$ for some $m\geq 1$, we have that $$\frac{5^n-1}{3} = \frac{5^{2m}-1}{3} = \frac{(5^m-1)(5^m+1)}{3}.$$For $m\geq 1$, $5^m-1$ and $5^m+1$ are both greater than $3$, so the expression cannot be prime, since if $3\mid 5^m-1$, then the expression is $\left(\frac{5^m-1}{3}\right)(5^m+1)$ which is the product of two integers greater than $1$, and if instead $3\mid 5^m+1$, then the expression is $\left(\frac{5^m+1}{3}\right)(5^m-1)$ which is again the product of two integers greater than $1$. Thus, for $n\neq 0$, $\frac{5^n-1}{3}$ cannot be a perfect square or a prime, and for $n = 0$, $\frac{5^n-1}{3}$ is a perfect square, so the answer is $n=0$ only, as claimed. $\blacksquare$

$\boxed{\textbf{CLAIM}}$ : No such n exists such that the $\frac{5^n-1}{3}$ prime. \[ \]$\textit{Proof} : $ Say there is a $n$ \[ 5^n - 1 \equiv 3p \pmod{4} \]\[ p \equiv 0 \pmod{4} \]$\boxed{\textbf{CLAIM}}$ $n=0$ is the only one integer such that $\frac{5^n-1}{3}=x^2$ \[ \]$\textit{Proof} : $ Assume $n \geqslant 1$ then \[ 5^n-1 \equiv 3x^2 \pmod{5} \]\[ -1 \equiv 3x^2 \pmod{5} \]Thats leads us with no sol for $n \geqslant 1$ as $x^2 \in \{0,1,-1\} \pmod{5}$. Thus, Our $\textbf{CLAIM}$ follows.

We need an even $n$ to ensure that $f(n)=\displaystyle{\frac{5^n-1}{3}}$ is an integer; therefore, we need $25^n-1$ to either be of the form $3a^2$ for a natural number $a$ or of the form $3p$ for a prime number $p$. In the first case, we get that $3a^2+1=25^n \implies a^2 \equiv 8 (\bmod \; 25)$, but $8$ is a quadratic non-residue $modulo \; 25$ - contradiction! In the second case, we know that $5^n-1=(5-1)(1+5+25+125+\dots+5^{n-1})$, so it would be always be divisible by $4$, thus making it not a prime. $\therefore$ No such $n$ exists. (other than $n=0$, for which quadratic residues do not really work)