Diophantine equation $1+5^k=2^y+2^z\cdot 5^t$ Find all positive integers solutions of the equation $1+5^k=2^y+2^z\cdot 5^t$.
My idea is, let $y,z\geq 2$. Then $4\mid RHS$. But $LHS\equiv 2(mod4)$ which is contradiction.
Let $y=1$. Then the equation becomes $1+5^k=2+2^z\cdot 5^t\implies 5^k-2^z\cdot5^t=1$ which is false by $(mod5)$.
And I don't know how to continue this solution if $z=1$. I know that $(k,y,z,t)=(2,4,1,1)$ is a solution and I'm trying to prove $k\leq2$. Any hint on this problem?
 A: In case of $z = 1$ I will prove that $k$ and $t$ cannot be both greater than $1$ (the remaining cases of $k = 1$ and $t = 1$ can be finished easily).  
In this case we get that $25$ divides $2^{y} - 1$. For that to be true $y$ has to be divisible by $4$ as the order of $2$ mod $5$ is $4$. Therefore $y = 4x$ and then:
$$25 \mid 2^{y} - 1 = 2^{4x} - 1 = 16^{x} - 1$$
By Lifting the Exponent Lemma we get that:
$$2 \le v_{5}(16^{x} - 1) = v_{5}(16 - 1) + v_5(x) = 1 + v_5(x)$$
Where $v_{p}(x)$ is the highest power of $p$ that divides $x$. So we get that $v_{5}(x) \ge 1$ which is equivalent to $5 \mid x$. Therefore $x = 5w$ and we get:
$$5^{k} - 2 \cdot 5^{t} = 2^{y} - 1 = 16^{x} - 1 = 16^{5w} - 1 = 1048576^{w} - 1 = (1048576 - 1)(...) = 1048575(...)$$
$104857 = 3 \cdot 5^2 \cdot 11 \cdot 31 \cdot 41$ so we get that all of these five primes divide:
$$5^{k} - 2 \cdot 5^{t} = 5^{t}(5^{k - t} - 2)$$
And therefore $3$, $11$, $31$ and $41$ must divide $5^{k - t} - 2$. $3$ and $41$ are of no use but in fact $11$ and $31$ give us the contradiction as no power of five is $2$ mod these primes. $11$ is smaller but $31$ is easier to check by hand as $5^{3} = 125 = 4 \cdot 32 + 1$ so the order of $5$ mod $31$ is just three and we have to eliminate just the first two powers.
A: COMMENT.-$1+5^k=2^y+2^z\cdot 5^t\Rightarrow6\equiv2^y\pmod{10}$  then $y=4n$ with $n\ge1$. It follows
$$1+5^k=16^n+2^z\cdot5^t$$
►$n=1\iff y=4\Rightarrow 5^{k-1}=3+2^z\cdot5^{t-1}\Rightarrow k=2,z=1,t=1$ so 
$(k,y,z,t)=(2,4,1,1)$ is a solution.
►$n=2\iff y=8\Rightarrow5^{k-1}=51+2^z\cdot5^{t-1}$.Therefore if $t\ge2$ we have
$5\equiv1\pmod{10}$ so $t=1$ but in this case we have $1\equiv3\pmod4$, absurde.
►$n=3\iff y= 12\Rightarrow5^{k-1}=819+2^z\cdot5^{t-1}=3^2\cdot7\cdot13+2^z\cdot5^{t-1}$. Like above $t=1$ but in this case we would have $1\equiv3\pmod4$, absurde.
I stop here. This is not an answer, it is a comment. 
