I fixed all the errors, question answered:
4^2x = (2^2x)²
4^x = 2^2x
Replacing:
(2^2x)² - 17 (2^2x) + 16 = 0
Let's call the commom term "2^2x" as "y".
y² - 17y + 16 = 0
Now that it is just a simple quadratic equation, you can use the quadratic formula:
Δ = b² - 4ac
Δ = -17² - 4.1.16
Δ = 289 - 64
Δ = 225
x = -b ± √Δ / 2a
x = 17 ± 15 / 2
x' = 16
x" = 1
y = 16 and 1
2^2x = 16
2^2x = 2^4
2x = 4
x = 4/2
x = 2
2^2x = 1
2x = 0
x = 0
X = 0 and 2.