Example 23 (1993 Korean Mathematical Olympiad) Find all non-negative integers $n$, such that $2^{2^{n}}+5$ is a prime number.
When $n=0$,
$2^{2^{n}}+5=2^{2^{0}}+5=2^{1}+5=2+5=7$ is a prime number.
When $n>1$, since $2 \equiv-1(\bmod 3)$, so $2^{2^{n}}+5 \equiv(-1)^{2^{n}}+2=1+2 \equiv 0(\bmod 3)$, and $2^{2^{n}}+5>3$, thus $2^{2^{n}}+5$ is a multiple of 3, and not equal to 3, hence it must be a composite number. In summary, the only non-negative integer that satisfies the condition is $n=0$.