Case 1 : $z$ is non negative
Then $x,y$ will also be non negative
Sub Case 1 : $x=0$
Then $4^y+1=5^z$ .
As we know $v_5(4^y+1)=1+v_5(y)$ ... $(i)$
let $y=5^k.a$
Using $(i)$ $4^{5^k.a}+1=5^{k+1}$
that imply the only possible value of $k$ is $0$ , and $a=1$
So Solution for this case is $(x,y,z)=(1,0,1)$
Sub Case 2 : $x>0$
Then using $mod3,4$ respectively we can say $z,x$ are even .
Now let $z=2a , x=2b$
So $2^{2y}=(5^a)^2-(3^b)^2$
$(5^a+3^b)(5^a-3^b)=2^{2y}$
that imply $(5^a+3^b)=2^{2y-t}$ ...$(ii)$
$(5^a-3^b)=2^t$ ... $(iii)$
$(ii)+(iii)$
$2.5^a=2^{2y-t}+2^t$
so $t=1$. and $5^a=2^{2(y-1)}+1=4^{y-1}+1$ .
Following Sub Case 1 :$ a=1,y=2$
So solution for this case is $(x,y,z)=(2,2,2)$
Case 2 : $z$ is negative
There is no solution indeed (divisibility) .
