I'm interested in the following diophantine eqaution: $(5^n-1)/4=y^2$. 


It turns out that this is a special case of the Nagell-Ljunggren equation, where $x=5$ and $q=2$


It has been shown that for x=5 this has no solutions but I'm looking for an elementary solution of this special case.



Special Case of famous Equation
This kind of problem usually requires a little algebraic number theory. Joe Silverman sketches one possible approach in the comments. Here is another. Let's rewrite as
$$
(2y)^2-5^n=-1.
$$
If $n$ is even then the left-hand side is a difference of two squares, which quickly gives a contradiction. So write $n=2m+1$. Then
$$
(2y+5^m \sqrt{5})(2y-5^m \sqrt{5})=-1.
$$
Thus $2y+5^m \sqrt{5}$ is a unit in $\mathbb{Z}[(1+\sqrt{5})/2]$. A fundamental unit is $\epsilon=(-1+\sqrt{5})/2$. It follows that
$$
2y+5^m \sqrt{5}=\pm \epsilon^{t}.
$$
Conjugating 
$$
2y-5^m \sqrt{5}=\pm \mu^t, \qquad \mu=(-1+\sqrt{5})/2.
$$
Taking differences and dividing by $\pm \sqrt{5}$ we have
$$
\pm 2 \cdot 5^m= \frac{\epsilon^t-\mu^t}{\sqrt{5}}.
$$
The right-hand side is the $t$-th Fibonacci number. Thus the equation becomes
$$
F\_t=\pm 2 \cdot 5^m.
$$
Let's rule out the case $m \ge 1$, which leaves you with $F\_t= \pm 2$. If $m \ge 1$ then $F\_t \equiv 0 \pmod{10}$. Now write out the Fibonacci sequence modulo $10$ and convince yourself that this forces $t \equiv 0 \pmod{15}$. But then $F\_{15} \mid F\_t$. However, $F\_{15}=2 \cdot 5 \cdot 61$, which does not divide $2 \cdot 5^m$, giving a contradiction.


