How to prove that there is an infinite number of primitive pythagorean triples such as $b=a+1$ and $2 | a$?

Let Euclid's formula for generating Pythagorean triples be $F(m,k)$ where

$$A=m^2-k^2qquad B=2mkqquad C=m^2+k^2$$ This is a reversal of the A and B in your equations but it works the same. We can now use a formula that generates the Pell numbers needed for input to Euclid's formula which in turn generates $(B-A=pm1)$ triples in sequential order of size. By inspection we can see that, starting with $(1)$, there are infinite solutions.

$$m=k+sqrt{2k^2+(-1)^k}$$ For example(s) begin{align*} k=1:qquad & m=(1+sqrt{2(1)^2+(-1)^1}space)big)=2qquad qquad & F(2,1)=(3,4,5)\ k=2:qquad & m=(2+sqrt{2(2)^2+(-1)^2}space)big)=5qquad qquad & F(5,2)=(21,20,29)\ k=5:qquad & m=(5+sqrt{2(5)^2+(-1)^5}space)big)=12qquad qquad & F(12,5)=(119,120,169)\ k=12:qquad & m=(12+sqrt{2(12)^2+(-1)^{12}}space)big)=29qquad qquad & F(29,12)=(697,696,985) end{align*}

Only the first version should be considered because $2|a$.In that case $2t^2+1= (s-t)^2$, Replace $s-t$ by $x$ and get a Pell equation $x^2-2t^2=1$ which has infinitely many solutions.This gives you infinitely many pairs $(s, t)$ and infinitely many pythagorean triples.