Can a Pythagorean Triple ever have TWO identical exponents > 1? For example, the triple
$(12, 16, 20)$
has these exponents:
$(12^1, 4^2, 20^1)$ or $(12^1, 2^4, 20^1)$
denoted by
$(1,2,1)$ or $(1,4,1)$

Here are more triples (ignore simple (1,1,1) cases):

I ran millions of triple tests looking for two identical exponents > 1, but found no matches.
Is there an existing proof or can someone prove it?
Thanks!
 A: EDIT: The original answer implicitly made the assumption that the sides of the triangle were relatively prime, which is not part of the original post. I've edited to explain why this assumption is valid.

Your question is equivalent to seeking an integer solution to $a^x+b^y=c^z$ in positive integers $a,b,c,x,y,z$, where $x,y,z$ are even and either $x=y\ge 4$ or $y=z\ge 4$. Note that since any $2k$th power is also a square, we can assume WLOG that the third exponent in either case is $2$.
First, let's see that we can reduce to the case where $a,b,c$ are relatively prime. Suppose that our bases have a common divisor $d>1$, and write them as $da,db,dc$ where $a,b,c$ are relatively prime. Then if $x=y$, we have
$$d^x(a^x+b^x)=d^2c^2\implies a^x+b^x=\left(\frac{c}{d^{x/2-1}}\right)^2$$
(Recall that $x$ is even and at least $4$, so $x/2-1$ is a positive integer.) So then $a^x+b^x$, an integer, is equal to the square of the rational number $c/(d^{x/2-1})$. But since the square of a rational number is an integer only when the original rational number is, the tuple $(a,b,c/(d^{x/2-1}))$ is another solution, but this time without any common factors.
A very similar analysis shows that in the case $y=z$, we can reduce to $(a/(d^{z/2-1}),b,c)$ and obtain a solution with relatively prime $a,b,c$. (This sort of reduction does not work if all three exponents are different; if it did, Oscar Lanzi's answer could be reduced to yield a solution with relatively prime $a,b,c$ and falsify Beal's conjecture.)
Having made this reduction, we can find several results in the literature tackling sums of powers of relatively prime positive integers.
The case $x=y$ has been proven impossible by Darmon and Merel (1995); see this citation in Wikipedia.
The case $y=z$ can be reduced to $z=2p$ for some prime $p$, since an $n$th power is an $m$th power if $m$ divides $n$. (Hence, we can ignore $k=8,12,16,18,20,24,28,30,\ldots$.)
$z=4$ follows from a relatively elementary proof by infinite descent, and dates to Fermat (see here for a proof).
$z=6,10$ follow from a result of Bennett, Chen, Dahmen and Yazdani (from Wikipedia's list of partial results on the Beal conjecture.)
As far as I am aware, $z\ge 14$ remains open (and is likely very difficult); one should at least note that the Fermat-Catalan conjecture implies there are at most finitely many solutions, and the ten known solutions to the Fermat-Calatan equation do not have even exponents, so at least no examples are known (very likely none exist).
A: We do know that all three exponents can assume different values greater than $1$.  Here a solution is constructed from the $(3,4,5)$ triple.
Multiply that triple by $9$ to get $(27,36,45)$.  This has one leg with exponent $2$ and the other leg with exponent $3$.  The hypoteneuse has no exponent greater than $1$, but now we take advantage of the fact that multiplying the $(27,36,45)$ triple by any sixth power preserves the exponents of $2$ and $3$ in the legs.  We choose $45^6$ as the multiplier and arrive at
$(6075^\color{blue}{3}, 546750^\color{blue}{2},45^\color{blue}{7}).$
