Solve over positive integers: $1+4^x=5^y$ Find all solutions to $1+4^\mathcal{x}=5^{\mathcal{y}}$ given that $x,y \in \mathbb{Z^{+}}$
Do give a proof.
Some calculus proof will be better. Non-calculus will also be okay. But since I am learning calculus, I would like that more.. 
I got no idea using calculus to be true..
Thanks in advance.
 A: Hints: Suppose $x>1$. Looking modulo $8$, note that $y$ must be even, say $y=2k$. Then $4^x=5^y-1=(5^k+1)(5^k-1)$. Now look at prime factorizations.
A: An elementary proof. You have
$$
5^{y} - 1 = (5 - 1) (1 + 5 + \dots + 5^{y-1}),
$$
so
$$
4^{x-1} =  1 + 5 + 5^{2} + \dots + 5^{y-1}.
$$
If $x > 1$, RHS is divisible by $4$. Since $5 \equiv 1 \pmod{4}$, we have that $y$ is divisible by $4$.
But then $5^{y} - 1$ is divisible by $5^{4} - 1 = 624$, which is not a power of $2$. (Thanks to @Wojowu for the correction.)
A: There is an obvious solution: $x=y=1$. Catalan's conjecture (aka Mihailescu's theorem) states that there is no more.
A: One might note, by the application of fermat's little theorm, applied to integers, that the period of $2^n$ in base 5, has a period of $2^{(n-2)}$.  This means that that the first power of 5 that is one more than a multiple of 16 is $5^4-1$, but this is the product of the double-even $5^2-1$ and the double-odd $5^2+1$.  Since every power of $5^n-1$ that 16 divides, so does 13, the only example of $5^a=4^b+1$ is for $a=b=1$.
