= Solution
For a <simple continued fraction>, define
$$
p_{-2}=0,\ p_{-1}=1,\quad q_{-2}=1,\ q_{-1}=0,\qquad p_n=a_np_{n-1}+p_{n-2},\quad q_n=a_nq_{n-1}+q_{n-2}.
$$
Multiplying the matrices $\begin{pmatrix}a_j&1\\1&0\end{pmatrix}$ gives
$$
\begin{pmatrix}p_{n-1}&p_{n-2}\\q_{n-1}&q_{n-2}\end{pmatrix},\qquad [a_0,\ldots,a_{n-1},\beta]=\frac{\beta p_{n-1}+p_{n-2}}{\beta q_{n-1}+q_{n-2}}.
$$
Their determinants also give $p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}$. If the infinite tail is $\beta=[a_{n+1},a_{n+2},\ldots]>a_{n+1}$, the same formula implies
$$
\left|\theta-\frac{p_n}{q_n}\right|=\frac1{q_n(\beta q_n+q_{n-1})}<\frac1{q_nq_{n+1}}.
$$
The denominators of these <continued fraction convergents> tend to infinity, since $q_{n+2}\geq q_{n+1}+q_n$. Hence they converge to $\theta$.
The square-root algorithm for the <periodic continued fraction of a quadratic irrational>, using $M'=Da-M$, $D'=(53-M'^2)/D$ and $a'=\lfloor(7+M')/D'\rfloor$, gives
$$
\sqrt{53}=[7;\overline{3,1,1,3,14}].
$$
The first four <continued fraction convergents> are $7/1,22/3,29/4,51/7$. Direct substitution gives \b[positive solutions to the requested Pell-type equations]:
$$
\boxed{51^2-53\cdot7^2=4,\qquad29^2-53\cdot4^2=-7.}
$$
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