Modulo 2, there are two possibilities for x, two possibilities for y, and two possibilities for z. If none of x, y, z are divisible by 2, then they are all congruent to 1 modulo 2. But
12 + 12 ≢ 12 mod 2,
so this is not a solution. Therefore at least one of them is divisible by 2.

A second way to show this is to let X = x2, Y = y2, and Z = z2. Then let's consider the congruence

X + Y ≡ Z mod 2.
Since 02 ≡ 0 mod 2 and 12 ≡ 1 mod 2, the possibilities for X, Y, and Z modulo 2 are also 0 and 1. Then test all eight possible combinations:
0 + 0 ≡ 0 mod 2
0 + 0 ≢ 1 mod 2
0 + 1 ≢ 0 mod 2
0 + 1 ≡ 1 mod 2
1 + 0 ≢ 0 mod 2
1 + 0 ≡ 1 mod 2
1 + 1 ≡ 0 mod 2
1 + 1 ≢ 1 mod 2
There are four non-solutions. The four solutions are
0 + 0 ≡ 0 mod 2
0 + 1 ≡ 1 mod 2
1 + 0 ≡ 1 mod 2
1 + 1 ≡ 0 mod 2.
Each solution contains at least one 0, which corresponds to one of X, Y, Z being divisible by 2, which corresponds to one of x, y, z being divisible by 2.

The second method is a little longer than the first but it generalizes more easily. Modulo 3, the possibilities for X (and also Y and Z) are

02 ≡ 0 mod 3,
12 ≡ 1 mod 3,
22 ≡ 1 mod 3.
So there are only two possibilies — 0 and 1 modulo 3. (The values 0 and 1 are called quadratic residues modulo 3, whereas 2 is a quadratic nonresidue modulo 3 because the square of an integer is never 2 modulo 3.) The only solutions to X + Y ≡ Z mod 3 are
0 + 0 ≡ 0 mod 3
0 + 1 ≡ 1 mod 3
1 + 0 ≡ 1 mod 3.
Each solution contains a 0, which corresponds to x, y, or z being divisible by 3.

Modulo 5 we play the same game:

02 ≡ 0 mod 5
12 ≡ 1 mod 5
22 ≡ 4 mod 5
32 ≡ 4 mod 5
42 ≡ 1 mod 5.
So there are only three possibilities for X (and also Y and Z) modulo 5. (The quadratic residues modulo 5 are 0, 1, and 4.) There are seven solutions to X + Y ≡ Z mod 5 with these values:
0 + 0 ≡ 0 mod 5
0 + 1 ≡ 1 mod 5
0 + 4 ≡ 4 mod 5
1 + 0 ≡ 1 mod 5
1 + 4 ≡ 0 mod 5
4 + 0 ≡ 4 mod 5
4 + 1 ≡ 0 mod 5.
(You can determine these solutions by hand, but it's more fun to write a short program to do it.) Each solution contains a 0, which corresponds to x, y, or z being divisible by 5.

Incidentally, you can use the same approach to show that if x2 + y2 = z2 then one of x, y is divisible by 4. And if x, y, z are relatively prime then z ≡ 1 mod 4. See the Pythagorean Triples Project for more.