finding infinetely many primes in three different forms of prime numbers below

The proof for $4k+3$ and $6k+5$ are small variants of the usual "Euclid" proof that there are infinitely many primes.

To show that there is a prime of the form $4k+3$ that is $\gt n$, we consider the number $N=4n!-1$. Not all prime factors of $N$ can be congruent to $1$ modulo $4$, and all prime factors of $N$ are $\gt n$.

The argument for $6k+5$ is essentially the same, we use $N=6n!-1$.

The argument for $4k+1$ is harder. Let $N=(2n!)^2+1$. We then use the fact that any odd prime divisor of a number of the form $x^2+1$ must be of the shape $4k+1$. For suppose to the contrary that $p$ divides $x^2+1$, where $p$ is of the form $4k+3$. Then $x^2\equiv -1\pmod{p}$. But it is an early result in the theory of quadratic residues that the congruence $x^2\equiv -1\pmod{p}$ has no solutions if $p$ is a prime of the form $4k+3$.

For $4k+3$ and $6k+5$ we can have this rather elementary proof.

Suppose that $p_1,...,p_n$ are all primes of the form $4k+3$ and hence finite. Then consider $m=4p_1....p_n+3$. If $m$ is not prime then all divisors of $m$ are of the form $4k+1$. But then $m\equiv 1 \mod 4$ which is not true. Therefore $m$ should have another divisor of the form $4k+3$ which is not $p_1,...,p_n$ and hence the primes of the form $4k+3$ are infinite.

The same idea works for $6k+5$ knowing that the odd primes, apart are of the form $6k+5$ and $6k+1$ and $3$. Therefore for $p_1,...,p_n$, all primes of the form $6k+5$, $m=4p_1....p_n+3$ should have another divisor of the form $6k+5$, otherwise $m\equiv 1\text{ or }3\mod 6$.