I compute below
$\displaystyle\begin{pmatrix}p_{n+1 } & p_{n } \\ q_{n+1 } & q_{n}\end{pmatrix}=\begin{pmatrix}2 & 1 \\ 1 & 0\end{pmatrix}^{n+1}.$
Notation: Instead of $p_{n+1}$ and $q_{n+1}$ let me use respectivelly $A_n$ and $B_n$.
With the initial conditions $A_{-1}=1$, $B_{-1}=0$, $A_{0}=b_{0}$, $B_{0}=1$,
$\displaystyle\begin{pmatrix}A_{\nu } & A_{\nu -1} \\ B_{\nu } & B_{\nu -1}\end{pmatrix}=\displaystyle\prod_{k=0}^{\nu }\begin{pmatrix}b_{k} & 1 \\ a_{k} & 0\end{pmatrix}\qquad(\text{with }a_0=1)\qquad(\ast)$
is the matrix notation of the truncated generalized continued fraction
$\dfrac{A_{n}}{B_{n}}=b_{0}+\underset{\nu =1}{\overset{n}{\mathbf{K}}}\left( \dfrac{a_{\nu}}{b_{\nu}}\right)=b_{0}+\dfrac{a_1}{b_1\ +}\ \dfrac{a_2}{b_2\ +\cdots}\ \ \dfrac{a_n}{b_n} \qquad (0)$
(the first expression is the Gauss Notation ). Its fundamental second order recurrence relation is given by
$\left\{ \begin{array}{c}b_{\nu }A_{\nu -1}+a_{\nu }A_{\nu -2}=A_{\nu } \\ b_{\nu }B_{\nu -1}+a_{\nu }B_{\nu -2}=B_{\nu }\end{array}\right. \qquad(\nu=1,2,\dots,n).\qquad (1)$
For $a_k=1$, we have respectivelly
$\displaystyle\begin{pmatrix}A_{\nu } & A_{\nu -1} \\ B_{\nu } & B_{\nu -1}\end{pmatrix}=\displaystyle\prod_{k=0}^{\nu }\begin{pmatrix}b_{k} & 1 \\ 1 & 0\end{pmatrix}\qquad(\text{with }a_0=1)\qquad(\ast\ast)$
and
$\dfrac{A_{n}}{B_{n}}=b_{0}+\underset{\nu =1}{\overset{n}{\mathbf{K}}}\left( \dfrac{1}{b_{\nu}}\right)\qquad (2)$
From the equality
$\sqrt{2}-1=\dfrac{1}{2+\sqrt{2}-1}=\dfrac{1}{2+\dfrac{1}{2+\sqrt{2}-1}}=\cdots $
we get the following expansion for $\sqrt{2}$
$\sqrt{2}=1+\underset{\nu =1}{\overset{\infty}{\mathbf{K}}}\left( \dfrac{1}{2}\right) =1+\dfrac{1}{2\ +}\ \dfrac{1}{2\ +}\ \ \dfrac{1}{2\ +\cdots}\qquad (3)$
whose convergents $A_n$ and $B_n$ are such that
$\dfrac{A_{n}}{B_{n}}=1+\underset{\nu =1}{\overset{n}{\mathbf{K}}}\left( \dfrac{1}{2}\right)\qquad (4)$
Since $a_k=1,b_k=2 (k\ge 1)$, the fundamental recurrence for $\sqrt{2}$ is given by the following system
$\left\{ \begin{array}{c}2A_{\nu -1}+A_{\nu -2}=A_{\nu } \\ 2B_{\nu -1}+B_{\nu -2}=B_{\nu }\end{array}\right. \qquad(\nu=1,2,\dots,n).\qquad (5)$
or this product of matrices
$\displaystyle\begin{pmatrix}A_{\nu } & A_{\nu -1} \\ B_{\nu } & B_{\nu -1}\end{pmatrix}=\displaystyle\prod_{k=0}^{\nu }\begin{pmatrix}2 & 1 \\ 1 & 0\end{pmatrix}=\begin{pmatrix}2 & 1 \\ 1 & 0\end{pmatrix}^{\nu}\qquad(\nu=1,2,\dots,n)\qquad(\ast\ast\ast)$
or in your notation, for $n=0,1,2,\dots$
$\displaystyle\begin{pmatrix}p_{n+1 } & p_{n } \\ q_{n+1 } & q_{n}\end{pmatrix}=\begin{pmatrix}2 & 1 \\ 1 & 0\end{pmatrix}^{n+1}$
This is the most compact form I am aware of. Hope this helps!
Generalization: For $\sqrt{1+x}$ one can use the algebraic identity
$\sqrt{1+x}-1=\dfrac{x}{2+\sqrt{1+x}-1}=\dfrac{x}{2+\dfrac{x}{2+\sqrt{1+x}-1}}=\cdots $
References
Oskar Perron, Die Lehre von den Kettenbrüchen, 3.ª ed., vol. II, B. G. Teubner, Stuttgart, 1957.
Lisa Lorentzen and Haakon Waadeland, Continued Fractions and Applications, North-Holland, Amsterdam, 1992.
Sergey Khrushchev , Orthogonal Polynomials and Continued Fractions: From Euler's Point of View, Cambridge University Press, 2008.