<s>[INST] Usage of Zsigmondy's theorem 
Find the solutions for the equation $$3^x-5^y=z^2$$ where $x,y,z$ are positive integers.

We find one solution $(x,y,z)=(2,1,2)$ by checking small cases. Initially looking at this modulo $4$ one gets that $$(-1)^x-1^y \equiv 0,1 \pmod 4$$ as squares leave residues $0$ and $1$ modulo $4$. This in turn implies that $x$ must be even that is $x=2k$ for integer $k$.
We then have that $$\begin{align}3^{2k}-z^2&=5^y \\ (3^k-z)(3^k+z) &= 5^y \end{align}$$ thus $$\begin{align} 3^k-z &= 5^a \\ 3^k+z &= 5^b \end{align}$$ for such that $b>a\ge1$ and $a+b=y$.
Addin these two equations we eliminate $z$ and get that $$2\cdot 3^k=5^a+5^b =5^a(5^{b-a}+1)$$ the lhs is a multiple of $2$ and the rhs is a multiple of $5$ this makes $5^a=1$ and $a=0$.
Now apparently this is the place where I could apply Zsigmondy's theorem, but I don't know how. Could someone englighten me how to use this theorem? It states that

Let $a>b\geq1$ be coprime integers, and let $n\geq 2$. Then there exists a prime divisor of $a^n-b^n$ that does not divide $a^k-b^k$ for all $1\leq k< n$, except when: $n=2$, and $a+b$ is a power of $2$; or $(a,b,n)=(2,1,6).$

but I don't see how to put the pieces together.
Edit: I think that since $$2\cdot 3^k=5^a+5^b =5^a(5^{b-a}+1)$$ and the fact that $a$ must be $0$ we get the result $b=y$ from $a+b=y$ and so $$2 \cdot 3^k = 5^y +1$$ now due to $\gcd(5,1) =1$ and $5>1$ we can use Zsigmondy's for $y \ge 2$ and we have the existence of prime $p$ for which $p \mid 5^y +1^y$, but $p \nmid 5+1$ i.e $p \nmid 6$. I still don't see how this implies the result?
 [/INST] There's 2 forms of Zsigmondy's theorem, one for forms $a^n-b^n$ and one for forms $a^n+b^n$. Your RHS is in the second form.
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