Conjecture: $xa^y+yb^z=zc^w+wd^x$ has no solutions in distinct positive integers satisfying $a + b + c + d = x + y + z + w$

Motivation:

While recently researching equations similar to this, I suddenly had an idea: would adding coefficients and specific restrictions change the difficulty? Therefore, I unexpectedly pieced together the following Diophantine equation:

Let $a, b, c, d, x, y, z, w$ be 8 pairwise distinct positive integers, satisfying:

$$a + b + c + d = x + y + z + w$$

Conjecture: The following exponential equation has no positive integer solutions:

$$x \cdot a^y + y \cdot b^z = z \cdot c^w + w \cdot d^x$$

I wrote a pythoncompiler to search for counterexamples. No valid 8-tuple can be found within the range $1 \le a, b, c, d, x, y, z, w \le 30$. (This is already the limit of my computer :D)

My question:

Can this conjecture be proven using existing theorems? Or are there counterexamples with larger values? Or are there any further restrictions on $a, b, c, d, x, y, z, w$ (for example, that they are all odd numbers)? Any answers are welcome!

Update (Based on @User's answer)

@User's counterexample has refuted the original conjecture.

As he suggested in his answer, if we further restrict all 8 variables to pairwise distinct positive odd numbers:

$$\text{Does } x \cdot a^y + y \cdot b^z = z \cdot c^w + w \cdot d^x \text{ have any solutions when } a,b,c,d,x,y,z,w \text{ are all odd?}$$

Since for odd numbers $u^n \equiv u \pmod 8$, the modulo $8$ operation does not rule out the existence of a solution. I welcome any insights into this case containing only odd numbers!

Update 2

Since @User demonstrated that an infinite parametric family of counterexamples exists even under the all-odd constraint.

Refined Question:

Let $a, b, c, d, x, y, z, w$ be 8 pairwise distinct PRIME numbers satisfying:

$$a + b + c + d = x + y + z + w$$

Does the Diophantine equation

$$x \cdot a^y + y \cdot b^z = z \cdot c^w + w \cdot d^x$$ have any finite solutions?