Prove the nonexistence of a Pythagorean triple for certain conditions

If for some natural numbers a, b, and c, a^2 + b^2 = c^2,
prove that (a+2)^2 + b^2 does not equal the square of a natural number.
[In other words, if a Pythagorean triple exists for natural numbers a, b, and c, there exists no Pythagorean triple with a+2, b, and another natural number]

(a+2)^2 + b^2 = a^2+b^2 +4a+4 = c^2 +4a+4

Suppose that a natural number like m exists such that c^2+4a+4=m^2

We know that m>c (because m^2>c^2)
We also know that a c^2+4a+4 or
c^2+4a+4<(c+2)^2
or
m^2<(c+2)^2
or m
so c
The only possibility is therefore m=c+1

c^2+4a+4=(c+1)^2
==>c^2+4a+4=c^2+2c+1
==> 4a+4 = 2c+1

This never happens because the LHS is even and the RHS is odd.

I think you miswrote this - as is, it's incorrect. For example, let a=1 and b=4:

(a+2)^2 + b^2 = (3^2) + (4^2) = 25 = 5^2

Counterexample, so it's false.