The answer is $p=2,3,7,19$. For $p=2$, use $z^{19}\equiv z\bmod{2}$. For $p=3$, use $z^{19}\equiv z\bmod{3}$. For $p=7$, use $z^{19}\equiv z\bmod{7}$. For $p=19$, use $z^{19}\equiv z\bmod{19}$. To see that no other prime $p$ works, note that $2^{19}-2=524286= 2\cdot3^3\cdot7\cdot19\cdot73$, so that the only remaining candidate is $p=73$. But $3^{19}\equiv70\bmod{73}$ and $2^{19}\equiv2\bmod{73}$, so that for $x=2$ and $y=1$ we have $(x + y)^{19} - x^{19} - y^{19}\equiv 3^{19} - 2^{19} - 1^{19}\equiv 67\bmod{73}$. Hence $p=73$ does not work.

This problem was proposed by Bulgaria

Let $x=y=1$; the the expression in question becomes $2^{19}-2 = 2\cdot 3^3 \cdot 7 \cdot 19 \cdot 73$. Thus, $p\in \{2,3,7,19,73\}$. Since $2-1, 3-1, 7-1, 19-1$ all divide $19-1=18$, by Fermat's Little Theorem, if $p\in \{2,3,7,19\}$, then, \[(x+y)^{19} - x^{19} - y^{19} \equiv (x+y) - x - y \equiv 0 \pmod{p}.\]We now claim that $p=73$ does not work. To do this, let $x=2$, $y=3$. Note that $\text{ord}_{73}(2) = 9$. Also, observe that $3^4 \equiv 2^3 \pmod{73}$, and $5^4\equiv -2^5 \pmod{73}$. It is now easy to find that $2^{19}\equiv 2 \pmod{73}$, $3^{19}\equiv 70 \pmod{73}$, and $5^{19}\equiv 62 \pmod{73}$, which finishes.

It seems to Freshmen‘s dream .