By brute-force search, yes. I started by searching over all tuples $(r,s,t)$ less than $1000$, stopping at the first example I found:
$$\left(138 + \sqrt{320}\right)^{570} \approx 10^{1249.9041}$$
In order to find the smallest example, I used the following strategy to search all examples smaller than the previous best (which I'll call $x$). Since the inside of the parenthesis $(r+\sqrt{s})$ must be at least $100+\sqrt{100}=110$, we can find the largest exponent we need to search by taking the logarithm:
$$
x = (r+\sqrt{s})^{t} > (110)^{t} \\
\log x > \log\left(110^t\right) \\
t < \frac{\log x}{\log 110}
$$
Thus we can start our search with the exponent at $\lfloor \log x / \log 110 \rfloor$ and search all exponents down to $100$.
For a given exponent $t$, we then need a way to search all pairs $(r,s)$. To put this mathematically:
$$
(r+\sqrt{s})^t < x \\
r + \sqrt{s} < x^{1/t}
$$
Starting with $r$, we start our search at $r=100$ and end it at:
$$
r + \sqrt{100} < x^{1/t} \\
r = \left\lfloor x^{1/t} - 10\right\rfloor
$$
Then, for each $r$ we search over $s$ from $s=100$ to:
$$
r + \sqrt{s} < x^{1/t} \\
s = \left\lfloor \left(x^{1/t} - r\right)^2\right\rfloor
$$
Every time we find a tuple $(r,s,t)$ with the desired property, we set $x$ to that new value (to reduce the number of tuples searched on further iterations). Note that any exponents we've already searched don't need to be searched again with the smaller limit.
Programmatically, this looks something like the following:
x = (138 + Sqrt[320])^570
For[t = Floor[Log[110, x]], t >= 100, t--,
For[r = 100, r < x^(1/t) - 10, r++,
For[s = 100, s < (x^(1/t) - r)^2, s++,
If[First@RealDigits[(r + Sqrt[s])^t, 10, 8, 3] == {2, 0, 1, 5, 2, 0, 1, 6},
x = (r + Sqrt[s])^t
]
]
]
]
(My actual code is a little different, but the strategy is the same.)
After a letting my search run for about a day, I realized that the search space was shrinking enough with each new $x$ that I could determine the smallest value in a reasonable timeframe. I let the search run for three more days, and the smallest value I found was:
$$
\left(140+\sqrt{27\ 027}\right)^{102} \\
\small \begin{split}
= \phantom{}
& 20473626230261221907121652389205011207101608771896002935841545009658234973292645 \\
& 17085491200709344560082101956950241075706483286182828328461907335813082258790620 \\
& 99769785021312880153909437562183881164238660035585722314971992368862943216420220 \\
& 1161896814\underline{2015.2016}08038424667587735557301174134230740327977685729631456117821\ldots
\end{split} \\
\approx 10^{253.311194\ldots}
$$
Rerunning the search produced no results, so I'm fairly confidant that this is the smallest example. There should be about 20 million smaller $(r,s,t)$, a fairly reasonable number to check by computer, so I would appreciate if someone could double-check my results!