For the equation

$$\huge (x^2-7x+11)^{(x^2-13x+42)}=1$$

Let's consider the following cases.

Case 1:

$$ \large 1^y = 1 \quad \implies \quad x=2,5$$

Case 2:

$$ \large \underbrace{y^0}_{y \neq 0} = 1 \quad \implies \quad x=6,7$$

Case 3:

$$ \large \underbrace{(-1)^{2n}}_{n \in \mathbb{Z}} = 1 \quad \implies \quad x=3,4$$

For the last case, we need to check the that the (exponent) power is even when $x=3$,

$$3^2-13 \times (3)+42=12$$

and when $x=4$,

$$4^2-13 \times (4)+42=6$$

which works!

Final answer:

$$ x=2,3,4,5,6,7$$

Checking with Mathematica:

Solve[(x^2 - 7*x + 11)^(x^2 - 13 x + 42) == 1, x, Reals]

>> {{x -> 2}, {x -> 3}, {x -> 4}, {x -> 5}, {x -> 6}, {x -> 7}}

For

$$\huge (x^2-7x+11)^{(x^2+13x-42)}=1$$

Similarly, we get

$$x=2,3,4,5$$ and

$$x=\frac{ -13 \pm \sqrt{337} }{2} $$

Using computer to check it, we get

Solve[(x^2 - 7*x + 11)^(x^2 + 13 x - 42) == 1, x, Reals]

>> {{x -> 2}, {x -> 3}, {x -> 4}, {x -> 5}, {x -> 1/2 (-13 - Sqrt[337])}, {x -> 1/2 (-13 + Sqrt[337])}}