Data type explanation regarding analogRead of a photoresistor
You should keep it uint16_t.
See: analogRead.
It shows the return value is a 10 bit number, which means a value of 0 to 1023.
If you use uint8_t, which is an 8 bit value, it only can store values from 0-255.
This means if the value is higher, the most significant bits are clipped/removed, and you only have the value module 256 left.
With a 16 bit value, you will have the left 6 bits unused, but this does not matter. It has the 10 bits you get from the analogRead function.
Example
Assume the value is 1000, which is in binary:
11 1110 1000

Written in 16 bits (using uint16_t), move it to the right and prefix with zeros you get:
0000 0011 1110 1000

However, if you try to store this in 8 bits (using uint8_t) you get:
(removed) 1110 1000

Which is the value 232 instead of 1000.
Some guidelines:

Always use a type that fits the value you want to store (in worst case).
If you know for sure your value will be non-negative, use an unsigned type (e.g. uint8_t).
If you want to preserve memory (or good practice) use a value that is not necessarily big (thus if you know a value will never be more than 1023 like in your case, don't use uint32_t).