Yes there are infinitely many. And it is not difficult to find them.
We seek continued fractions of the form
$\sqrt{N}=[a,\overline{b,c,2a}]$
First off, add $a$ to get a "pure" periodic expression. We shall call the quadratic surd $x$:
$x=a+\sqrt{N}=[\overline{2a,b,c}]$
We may then render
$x=2a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{x}}}$
and upon clearing fractions
$(bc+1)x^2+(b-c-2a(bc+1))x-(2ab+1)=0$
Now comes the sneaky part. If the above quadratic equation over the integers is to have a root $a+\sqrt{N}$, its other root must be $a-\sqrt{N}$ forcing the linear coefficient to be exactly $-2a$ times the quadratic one! Thereby $b=c$ above and the quadratic equation simplifies to:
$(b^2+1)x^2-2a(b^2+1)x-(2ab+1)=0$
This gives an integer radicand whenever $2ab+1$ is a multiple of $b^2+1$, in which cases the common factor of $b^2+1$ may be cancelled from the quadratic equation leaving the equation monic.
Suppose, for example, we drop in $b=2$. Then $2ab+1$ is to be a multiple of $5$ and $a$ can be any whole number one greater than a multiple of $5$. Putting $a=1$ results in the "trivial" solution $\sqrt{2}=[1,\overline{2}]$, as the period is reduced from three to one due to $b=c=2a$. But this equality is avoided for larger eligible values of $a$ and we get a series of period $3$ solutions. In all cases $N$ is one fourth the discriminant of the monic polynomial obtained after cancelling out the $b^2+1$ factor:
$a=6\to \sqrt{41}=[6,\overline{2,2,12}]$
$a=11\to \sqrt{130}=[11,\overline{2,2,22}]$
$a=5k+1\to \sqrt{25k^2+14k+2}=[5k+1,\overline{2,2,10k+2}]$
There are more families of solutions like this with other values of $b$. Just put in an even positive value for $b$ (why even?) and turn the crank. You must put $a>b/2$ to avoid the collapse we saw above with $\sqrt{2}$.
So much for a repeat petiod of $3$, what about larger periods?
Claim: For any positive whole numbers $r$ there are at least an infinitude of $\sqrt{N}$ continued fractions having repeat period $r$ where $N$ is a whole number, having the following form:
$\sqrt{N}=[kP_r+1;\overline{2,2,...,2,2(kP_r+1)}]$
$P_r$ is a Pell number defined by $P_0=0,P_1=P_{-1}=1,P_r=2P_{r-1}+P_{r-2}\text{ for } r\ge 2$, and $k$ is a whole number $\ge 0$ for $r=1$, $\ge 1$ otherwise. The number of $2$ digits before the final entries is $r-1$.
The proof bears some similarities to calculating the general solution for $r=3$ above. First add $kP_r+1$ to the expression to make a purely periodic fraction:
$x=kP_r+1+\sqrt{N}=[\overline{2(kP_r+1),2,2,...,2}]$
Then
$x=2(kP_r+1)+\dfrac{1}{[2,2,...,2,x]}$
By mathematical induction on $r$ and using the recursive relation defined for Pell numbers in the claim it is true that
$[2,2,...,2,x]=\dfrac{P_rx+P_{r-1}}{P_{r-1}x+P_{r-2}}$
with $r-1$ digits of $2$ in the block. When this is substituted into the previous equation the following is obtained:
$x=2(kP_r+1)+\dfrac{P_{r-1}x+P_{r-2}}{P_rx+P_{r-1}}$
$x=\dfrac{(2(kP_r+1)P_r+P_{r-1})x+2(kP_r+1)P_{r-1}+P_{r-2}}{P_rx+P_{r-1}}$
$(P_r)x^2-2(kP_r+1)P_rx-(2(kP_r+1)P_{r-1}+P_{r-2})=0$
Upon completing the square and back-substituting $\sqrt{N}=x-(P_rk+1)$ we obtain:
$N=\dfrac{(kP_r+1)^2P_r+2(kP_r+1)P_{r-1}+P_{r-2}}{P_r}$
Using the Pell number recursion to eliminate $P_{r-2}$:
$N=\dfrac{(kP_r+1)^2P_r+2(kP_r)P_{r-1}+P_r}{P_r}=(kP_r+1)^2+2kP_{r-1}+1$
thereby identifying $N$ as a whole number. For a full fundamental period $\ge 2$ the terminal element must not match the other elements, so in that case $k\ge 1$. Else (meaning a period of just $1$), $k$ may be any whole number, $k\ge 0$.
Best Answer
It is more convenient to look at this in terms of
$$\omega = \sqrt{n} + \lfloor \sqrt{n}\rfloor = \bigl[\overline{q_0,q_1,\dotsc,q_{l-1}}\bigr]\,.$$
It is clear that we have $q_0 = 2a_0$ and $q_k = a_k$ for $k \geqslant 1$, since $\omega$ differs from $\sqrt{n}$ exactly by the integer $a_0$.
Proposition: The partial quotients of the simple continued fraction expansion of $\omega$ satisfy $q_k \leqslant a_0$ for $k \not\equiv 0 \pmod{l}$, where $l$ is the minimal period of the continued fraction expansion. Additionally, $q_k = a_0$ can happen at most once per period, in the middle.
Proof: We can write the complete quotients of $\omega$ in the form $$\omega_k = \frac{\sqrt{n} + b_k}{c_k}$$ with integers $b_k, c_k$ satisfying
This is proved by induction. The base case $k = 0$ is immediate, for $$\omega_0 = \omega = \frac{\sqrt{n} + \lfloor \sqrt{n}\rfloor}{1}\,,$$
i.e. $b_0 = \lfloor \sqrt{n}\rfloor$ and $c_0 = 1$. For the induction step, we note that $b_{k+1} = q_{k}c_{k} - b_{k}$ and $c_{k+1} = \frac{n - b_{k+1}^2}{c_{k}}$. Since $b_{k+1} \equiv -b_k \pmod{c_k}$, we have $$(n - b_{k+1}^2) \equiv (n - b_{k}^2) \equiv 0 \pmod{c_{k}}$$ by the induction hypothesis, so $c_{k+1}$ is an integer (that $b_{k+1}$ is an integer is obvious). And by the continued fraction algorithm \begin{align} \omega_{k+1} &= \frac{1}{\omega_k - q_k} \\ &= \biggl(\frac{\sqrt{n} + b_k}{c_k} - q_k\biggr)^{-1} \\ &= \biggl(\frac{\sqrt{n} - (q_kc_k - b_k)}{c_k}\biggr)^{-1} \\ &= \frac{c_k}{\sqrt{n} - b_{k+1}} \\ &= \frac{c_k(\sqrt{n} + b_{k+1})}{n - b_{k+1}^2} \\ &= \frac{\sqrt{n} + b_{k+1}}{c_{k+1}}\,. \end{align}
Thus it is immediate that 3. holds for $k+1$ if it holds for $k$, and 2. follows from 1. since we always have $\omega_{k+1} > 1$. To see that 1. holds, note that $b_{k+1} < \sqrt{n}$ is equivalent to $\omega_k - q_k = \omega_k - \lfloor \omega_k\rfloor > 0$, which is true because $\omega_k$ is irrational. And \begin{align} && 0 &< b_{k+1} \\ &\iff& 0 &< q_k c_k - b_k \\ &\iff& \frac{b_k}{c_k} &< q_k\,. \end{align}
If $b_k < c_k$, then this is true because $q_k = \lfloor \omega_k\rfloor \geqslant 1$. And if $c_k \leqslant b_k$, then $c_k < \sqrt{n}$ and thus $$q_k = \biggl\lfloor \frac{\sqrt{n} + b_k}{c_k}\biggr\rfloor \geqslant \biggl\lfloor \frac{c_k + b_k}{c_k}\biggr\rfloor = 1 + \biggl\lfloor \frac{b_k}{c_k}\biggr\rfloor > \frac{b_k}{c_k}\,.$$
Having established $b_k \leqslant \lfloor\sqrt{n}\rfloor = a_0$ and $c_k > 0$ for all $k$, we see that
$$q_k = \biggl\lfloor \frac{\sqrt{n} + b_k}{c_k}\biggr\rfloor = \biggl\lfloor \frac{a_0 + b_k}{c_k}\biggr\rfloor \leqslant \biggl\lfloor \frac{2a_0}{c_k}\biggr\rfloor\,.$$
Thus $q_k \leqslant a_0$ unless $c_k = 1$. But if $c_k = 1$, then clearly $q_k = a_0 + b_k$, $b_{k+1} = a_0 = b_1$ and $c_{k+1} = n - a_0^2 = c_1$, i.e. $\omega_{k+1} = \omega_1$, which means $k$ is a multiple of the minimal period $l$.
Additionally, we see that $q_r = a_0$ if and only if $c_r = 2$ and $b_r = a_0$. Then we have $b_{r+1} = a_0 = b_r$ and $$c_rc_{r+1} = n - b_{r+1}^2 = n - b_r^2 = c_{r-1}c_r\,,$$ whence $c_{r+1} = c_{r-1}$. From this one finds $$b_{r+m} = b_{r+1-m} \qquad\text{and}\qquad c_{r+m} = c_{r-m}$$ as well as $q_{r+m} = q_{r-m}$ for $0 \leqslant m \leqslant r$ using the recursion $b_{k+1} = q_kc_k - b_k$ and $c_kc_{k+1} = n - b_{k+1}^2$. Thus if $q_r = a_0$, then $c_{2r} = c_0 = 1$ and $2r$ is a multiple of $l$. So $q_k = a_0$ can happen at most once in a period, and if it happens the period has even length and $q_{l/2} = a_0$.