By the Sophie Germain factorization, $19^{4n} + 4 = (19^{2n} - 2\cdot 19^n + 2)(19^{2n} + 2\cdot 19^n + 2) = ((19^n-1)^2 + 1) ((19^n+1)^2 + 1) $. The two bracket factors are obviously co-prime. Always one of them is divisible by $5$, but cannot be a power of $5$ (by Catalan). Therefore there are at least $3$ distinct prime divisors. And since for $n=1$ we have $19^4+4 = 5^2\cdot 13\cdot 401$, that minimum of $3$ is reached.

I have been asked to clarify my "by Catalan". Say $(19^n-1)^2 + 1 = 5^k$. The Catalan conjecture (proved by Preda Mihailescu) states that the only consecutive perfect powers are $8$ and $9$, so $(19^n-1)^2$ and $5^k$ cannot be. Similarly for the other case.

I think you missed something, you should prove also that for $n>1$: $19^{2n}+4$ is not of the form $5^kp^lq^m$ where $p+q$ is small than $401+13$?

Let's start with the idea in the solution given above: In the well known identity, $$ x^4 + 4 = (x^2 - 2x +2)(x^2 +2x +2) \qquad \dots (1) $$if $x$ is an odd number, then $(x^2 - 2x +2, x^2 +2x +2) = (x^2 - 2x +2, 4x) = (x^2 - 2x +2, x) = (2, x) = 1$. Hence, $x^2 + 2x +2$ and $x^2 - 2x +2$ are relatively coprime numbers. For$ x=19^n $, Let's take $A=19^{2n}+2\cdot 19^n +2$ and $B=19^{2n}-2\cdot 19^n +2$. Therefore $(1)$ identity will be $$ 19^{4n}+4 = (19^{2n}+2\cdot 19^n +2)(19^{2n}- 2\cdot 19^n +2) = A\cdot B \qquad \dots (2)$$ We conclude that $19^{4n} +4 $ has at least two prime divisors. $A\geq 401$ and $B \geq 325$. If $n$ is an even number, then $A \equiv 1 + 2\cdot (-1)^n +2 \equiv 0 \pmod{5}$ If $n$ is an odd number, then $B \equiv 1 + 2\cdot (-1)^n +2 \equiv 0 \pmod{5}$ If the number $ 19^{4n}+4 $ to have exactly two prime factors; either $A=5^m$, $B\neq 5$ is a prime or $B=5^m$, $A\neq 5$ ($m\in \mathbb Z^+$). Let's show that this is not possible. 1. Case: Let's solve $A=19^{2n}+2\cdot 19^n +2=5^m$ over positive integers. Let's examine the equation over $\mod {19}$, We find $5^m \equiv 2 \pmod{19}$. But: for $m=1,2,3,4,5,6,7,8,9$ respectively $5^m$ takes the values $$ 5^m \equiv 5 , 6 , 11 , 17 , 9 , 7 , 16 , 4 , 1 \pmod{19} \qquad \dots (3)$$. That is $5^m \not\equiv 2 \pmod{19}$. 2. Case: Similarly, let's solve $B=19^{2n}-2\cdot 19^n +2=5^m$ over positive integers. Let's examine the equation over $\mod {19}$, We find $5^m \equiv 2 \pmod{19}$. By $(3)$ $5^m \not\equiv 2 \pmod{19}$, is a contradiction. So, the number $19^{4n} +4$ has at least $3$ distinct prime divisors. For the example case, take $n=1$ and $19^4 +4=401\cdot 325 = 5^2\cdot 13 \cdot 401$ has $3$ distinct prime divisors.