$k \in Z^+$
firstly we know that there exists infinetly many primes of the form $4n+1$ by FTA also we see that if we consider finite primes say to $n$ then the recursive formular can be expressed as:
$p_k=4p_{k-1}+1=4^kp_0+4^{k-1}+4^{k-2}...+4^2+4+1=4^kp_0+\frac{4^{k}-1}{3}$
since $4\equiv 1 \mod 3$ we know there exists such an integral number $p_k$
On
There is no such sequence. For if there is, there is a sequence that starts at a number greater than $5$. We assume there is such a sequence, and derive a contradiction.
If $p$ is a term of the sequence, and is congruent to $1$ modulo $3$, then we must take $4p+1$ as the next term to avoid divisibility by $3$. We then get a number congruent to $2$ modulo $3$. Similarly, if we are at a number $q$ congruent to $2$ modulo $3$, we must go to $4q-1$, and we get a number congruent to $1$ modulo $3$.
So $\pm 1$ must alternate if we are to keep getting primes.
Now it is straightforward to verify that if we start with a number not divisible by $5$, then one of the next $4$ terms is divisible by $5$.
Your instinct of looking at divisibility by $3$ is a good one; you just need to look at it a from a slightly different angle. Firstly, note that since $4\equiv 1\bmod 3$, $p_k\equiv p_{k-1}\pm1\bmod 3$; this means that two consecutive $+1$ steps are impossible, since if $p\equiv 2\bmod 3$ then $4p+1\equiv 0\bmod 3$ so it can't possibly be prime, and likewise if $p\equiv 1\bmod 3$ then $4(4p+1)+1\equiv 0\bmod 3$. Similarly, there can't be two consecutive $-1$ steps for exactly the same reason. So assuming $p_0\equiv 1\bmod 3$ (if it's not, we can just 'start' on $p_1$ instead), we know that we must have $p_1=4p_0+1$, $p_2=4p_1-1$, $p_3=4p_2+1$, etc; the $\pm$ must alternate signs to avoid divisibility by 3. But this implies that $p_{2n+2}$ $= 4p_{2n+1}-1$ $= 4(4p_{2n}+1)-1$ $=16p_{2n}-3$. Now, consider $p_i \bmod 5$ and show that at least one of $p_{2n}, p_{2n+2}, p_{2n+4}, p_{2n+6}, p_{2n+8}$ must be divisible by $5$.