How many distinct positive integer-valued solutions exist to the equation ${(x^2-7x+11)}^{x^2-13x+42}=1$?

For an equation of the form $A^B=1$, there are three possible cases:

Case 1: The exponent is zero and the base is non-zero.
$x^2-13x+42=0$
$(x-6)(x-7)=0$
This gives $x=6$ and $x=7$.
Checking the base for these values:
For $x=6$: $6^2-7\left(6\right)+11=36-42+11=5\ne 0$.
For $x=7$: $7^2-7\left(7\right)+11=11\ne 0$.
So, $x=6,7$ are valid positive integer solutions.

Case 2: The base is equal to 1.
$x^2-7x+11=1$
$x^2-7x+10=0$
$(x-2)(x-5)=0$
This gives $x=2$ and $x=5$.
Since these are positive integers, they are valid solutions.

Case 3: The base is equal to -1 and the exponent is an even integer.
$x^2-7x+11=-1$
$x^2-7x+12=0$
$(x-3)(x-4)=0$
This gives $x=3$ and $x=4$.
Checking if the exponent is even for these values:
For $x=3$: Exponent = $3^2-13\left(3\right)+42=9-39+42=12$ (even).
For $x=4$: Exponent = $4^2-13\left(4\right)+42=16-52+42=6$ (even).
So, $x=3,4$ are also valid positive integer solutions.

Summarizing all valid solutions: $x\in \{2,3,4,5,6,7\}$.
Thus, there are 6 distinct positive integer solutions.