Here's \(\displaystyle 4n+1 \):

Again, assume there are only finitely many primes of the form \(\displaystyle 4n+1 \). Define \(\displaystyle N=(2p_1p_2\cdots p_k)^2+1 \), where \(\displaystyle \{p_1,\;p_2,\;\ldots,\;p_k\} \) is the set of the primes of the form \(\displaystyle 4n+1 \). Since \(\displaystyle N \) is of the form \(\displaystyle 4n+1 \), \(\displaystyle N>p_k \implies N \) is composite.

Let \(\displaystyle p\mid N \). Since \(\displaystyle N \) is odd, \(\displaystyle p\neq2 \). Thus \(\displaystyle (2p_1p_2\cdots p_k)^2+1\equiv0\bmod{p}\implies (2p_1p_2\cdots p_k)^2\equiv-1\bmod{p}\implies x^2\equiv-1\bmod{p} \) is solvable. By your post

here, we see that this implies \(\displaystyle p\equiv1\bmod{4}\implies p \) is of the form \(\displaystyle 4n+1 \).

Therefore \(\displaystyle p\in\{p_1,\;p_2,\;\ldots,\;p_k\}\implies p\mid(2p_1p_2\cdots p_k)^2 \). So \(\displaystyle p\mid N-(2p_1p_2\cdots p_k)^2=1 \) which is a contradiction. Using the same reasoning as in the above post, we see that there are an infinite amount of primes of the form \(\displaystyle 4n+1 \).