Inspired by tamirat solomon

Lets apply all values one by one
See only the last digits.
7^1 +6^1 = 5^1(See only last digits)
13=5(See last digits)
3=5 false
Again, See only the last digits.
7^2 + 6^2 = 5^2
49 + 36 = 25
In that also see the last digit.
9+6=15 So, this can be true. Like this see all others also
7^3 + 6^3=5^3
3 + 6 = 5 (only the lasts digits) So, this is false
7^4 +6^4 = 5^4
1 + 6 = 5(last digits) false
Thus the only possible value of n is 2 . So the answer is 2

Only squares of two integers can be equal to square of a third integer. Sum of any other power, n $\neq$ 2, of two integers can never be equal to n.th power of any integer.

We can write the above equation as:

$297^n + 396^n = 495^n \Rightarrow (3^3 11^1)^n + ( 2^2 3^2 11^1)^n = (3^2 5^1 11^1)^n \Rightarrow 3^n + 4^n = 5^n$

which is satisfied for $\boxed{n=2}.$

Just at sight remenbering Fermat Last Theorem, we must exclude exponent 3 and 4 then the only answer left is the sum of squares.

If we divide all terms by the largest common divisor, 99 we get
295/99=3 396/99=4 495/99=5

A triangle with sides equal to 3, 4, and 5 will define a right triangle. The Pythagorean Theorem gives us a^2 + b^2 = c^2.

n=2

This just a simple question