Solution:
We begin with the standard factorisation
$$y^{4}+4 = \left(y^{2}-2y+2\right)\left(y^{2}+2y+2\right)$$
Thus we have $y^{2}-2y+2 = p^{m}$ and $y^{2}+2y+2 = p^{n}$ for some positive integers $m$ and $n$ such that $m+n = x$. Since $y^{2}-2y+2 < y^{2}+2y+2$, we have $m < n$ so that $p^{m}$ divides $p^{n}$. Thus $y^{2}-2y+2$ divides $y^{2}+2y+2$. Writing $y^{2}+2y+2 = y^{2}-2y+2 + 4y$, we infer that $y^{2}-2y+2$ divides $4y$ and hence $y^{2}-2y+2$ divides $4y^{2}$. But
$$4y^{2} = 4\left(y^{2}-2y+2\right) + 8(y-1)$$
Thus $y^{2}-2y+2$ divides $8(y-1)$. Since $y^{2}-2y+2$ divides both $4y$ and $8(y-1)$, we conclude that it also divides $8$. This gives $y^{2}-2y+2 = 1, 2, 4$ or $8$.
If $y^{2}-2y+2 = 1$, then $y = 1$ and $y^{4}+4 = 5$, giving $p = 5$ and $x = 1$. If $y^{2}-2y+2 = 2$, then $y^{2}-2y = 0$ giving $y = 2$. But then $y^{4}+4 = 20$ is not the power of a prime. The equations $y^{2}-2y+2 = 4$ and $y^{2}-2y+2 = 8$ have no integer solutions. We conclude that $(p, x, y) = (5, 1, 1)$ is the only solution.

Alternatively, using $y^{2}-2y+2 = p^{m}$ and $y^{2}+2y+2 = p^{n}$, we may get
$$4y = p^{m}\left(p^{n-m}-1\right)$$
If $m > 0$, then $p$ divides $4$ or $y$. If $p$ divides $4$, then $p = 2$. If $p$ divides $y$, then $y^{2}-2y+2 = p^{m}$ shows that $p$ divides $2$ and hence $p = 2$. But then $2^{x} = y^{4}+4$, which shows that $y$ is even. Taking $y = 2z$, we get $2^{x-2} = 4z^{4}+1$. This implies that $z = 0$ and hence $y = 0$, which is a contradiction. Thus $m = 0$ and $y^{2}-2y+2 = 1$. This gives $y = 1$ and hence $p = 5, x = 1$.