Solution:
If $x = y = 1$ then $p$ divides
$$2^{19} - 2 = 2\left(2^{18} - 1\right) = 2\left(2^{9} - 1\right)\left(2^{9} + 1\right) = 2 \cdot 511 \cdot 513 = 2 \cdot 3^{3} \cdot 7 \cdot 19 \cdot 73$$
If $x = 2$, $y = 1$ then
$$p \mid 3^{19} - 2^{19} - 1$$
We will show that $3^{19} - 2^{19} - 1$ is not a multiple of $73$. Indeed,
$$3^{19} \equiv 3^{3} \cdot (3^{4})^{4} \equiv 3^{3} \cdot 8^{4} \equiv 3^{3} \cdot (-9)^{2} \equiv 27 \cdot 81 \equiv 27 \cdot 8 \equiv 70 \pmod{73}$$
and
$$2^{19} \equiv 2 \cdot 64^{3} \equiv 2 \cdot (-9)^{3} \equiv -18 \cdot 81 \equiv -18 \cdot 8 \equiv -144 \equiv 2 \pmod{73}$$
Thus $p$ can be only among $2, 3, 7, 19$. We will prove all these work.

- For $p = 19$ this follows by Fermat's Theorem as
$$(x+y)^{19} \equiv x + y \pmod{19}, \quad x^{19} \equiv x \pmod{19}, \quad y^{19} \equiv y \pmod{19}$$

- For $p = 7$, we have that
$$a^{19} \equiv a \pmod{7}$$
for every integer $a$. Indeed, if $7 \mid a$, it is trivial, while if $7 \nmid a$, then by Fermat's Theorem we have
$$7 \mid a^{6} - 1 \implies 7 \mid a^{18} - 1$$
therefore
$$7 \mid a(a^{18} - 1)$$

- For $p = 3$, we will prove that
$$b^{19} \equiv b \pmod{3}$$
Indeed, if $3 \mid b$, it is trivial, while if $3 \nmid b$, then by Fermat's Theorem we have
$$3 \mid b^{2} - 1 \implies 3 \mid b^{18} - 1$$
therefore
$$3 \mid b(b^{18} - 1)$$

- For $p = 2$ it is true, since among $x + y$, $x$ and $y$ there are $0$ or $2$ odd numbers.