Use divisibility of $3$

Given,

$p=3^w,q=3^x,r=3^y,s=3^z$ and $p^2+q^3+r^5=s^7$

So, $3^{2w}+3^{3x}+3^{5y}=3^{7z}$

Let, $a=2w,b=3x,c=5y,d=7z$

So $3^a+3^b+3^c=3^d$

Without loss of generality, we assume $a \leq b \leq c $

Then $3^a(1+3^{b-a}+3^{c-a})=3^d$

Now right hand side is a power of $3$.

So no other prime than $3$ divides left hand side

So $a=b=c$

So $3 \times 3^a=3^d$

$3^{a+1}=3^d$

So $a+1=d$

So $2w=3x=5y=7z-1$

To find the minimum value of $w+x+y+z$,

We have to find the minimum number $n$ such n is divisible by $2,3,5$ and $n+1$ is divisible by $7$

We find such $n$ is $90$.

So $w=45,x=30,y=18,z=13$

So, minimum $w+x+y+z=45+30+18+13=106$