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?

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. 
