Browsing around online, I find a handful of proofs that $3$ is always a primitive root of any Fermat prime $2^n+1$. One particular proof is found in problems 4 and 5 here. Another proof I found on a page at SUNYSB makes use of Pepin's test whose proof seems to require quadratic reciprocity.
So out of curiosity, is it possible to prove that any Fermat prime has $3$ as a primitive root without use of quadratic reciprocity?
The few observations I do know is that if $2^n+1$ is prime for $n\gt 1$, then $n$ is actually a power of $2$, and that in this case $U(\mathbb{Z}/p\mathbb{Z})$ has order $2^n$, so the order of $3$ must also be a power of $2$, but I couldn't get much further than that.