The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$

$x,x+1,z$ is a Pythagorean triple iff $(2x+1)^2+1=2z^2$.

Let $u=2x+1$. Then $u^2-2z^2=-1$, a negative Pell equation whose solution lies in considering the units of $\mathbb Z[\sqrt 2]$ of norm $-1$.

It is clear that $\omega=1+\sqrt 2$ is a fundamental unit with norm $-1$. Therefore, all the other solutions of $u^2-2z^2=-1$ come from odd powers of $\omega$.

Thus, if $(u_k,z_k)$ is a solution of $u^2-2z^2=-1$, then the next one is given by

$$\begin{align} u_{k+1}+z_{k+1}\sqrt 2&=(u_k+z_k\sqrt 2)\omega^2 \\ &=(u_k+z_k\sqrt 2)(3+2\sqrt 2) \\ &=(3u_k+4z_k)+(2u_k+3z_k)\sqrt 2 \end{align}$$ So, $u_{k+1}= 3u_k+4z_k$ and $z_{k+1}=2u_k+3z_k$.

Now let $u_{k+1}=2x_{k+1}+1$.

Then $$x_{k+1}=\frac{u_{k+1}-1}{2}=\frac{(3u_k+4z_k)-1}{2}=\frac{3(2x_k+1)+4z_k-1}{2}=3x_k+2z_k+1$$ and $$z_{k+1}=4x_k+3z_k+2$$ as claimed.

You are on the right track. The simplest solution is to recall that all irreducible Pythagorean triples for a rooted ternary tree beginning with $(3, 4, 5) $ triangle. B Berggren discovered that all others can be derived from this most primitive triple. F J M Barning set these out as three matrices that when pre-multiplied by a "vector" of a Pythagorean triple produces another. For the case of consecutive legs we have, starting with $(x_1, y_1, z_1) $, we may calculate the next triple as follows:

$$\begin {align} x_2&=x_1+2y_1+2z_1 \\ y_2&=2x_1+y_1+2z_1 \\ z_2&=2x_1+2y_1+3z_1 \end {align} $$

The hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof by induction to show that the formulas are correct. The first few are:

$(3, 4, 5); (20, 21, 29); (119, 120, 169); (696, 697, 985); (4059, 4060, 5741); (23660, 23661, 33461)$; etc. Obviously, this can be continued indefinitely.

The sequence rises geometrically. A simple explicit formula is available for these solutions that are (as you have already guessed) alternating solutions to Pell's equation.