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 $.