$1+\sqrt{2}$ seems special: its powers quickly approach integers $ \alpha = 1+\sqrt{2}$ has some interesting properties. One of which is that if you look at powers
\begin{align*}
\alpha^2 &= 3+2\sqrt{2} \approx 5.8284 \\
\alpha^3 & \approx 14.0710 \\
\alpha^4 & \approx 33.9705 \\
& \vdots \\
\alpha^{10} & \approx 6725.9998 \\
\alpha^{11} & \approx 16238.0000 \\
\vdots & 
\end{align*}
(values truncated, not rounded) then the sequence $\{ \alpha^n \}$ is "almost" an unbounded integer sequence. What is going on here? Is this property secondary to $\alpha$ being a unit in $\mathbb{Z}[\sqrt{2}]$? Can $\alpha$ be interpreted as a type of approximate eigenvalue? Does this behavior have to do with the continued fraction representation?
Other numbers such as $1+\sqrt{3}$ have a similar property, but they don't appear to "become integer" as quickly. 

You could observe the pattern if you would calculate the powers:
$$\begin{align}\alpha^1&=(1+\sqrt{2})^1=1+1\sqrt{2}\\
a^2&=(1+\sqrt{2})^2=3+2\sqrt{2}\\
a^3&=(1+\sqrt{2})^3=7+5\sqrt{2}\\
a^4&=(1+\sqrt{2})^4=17+12\sqrt{2}\\
\vdots
\end{align}$$
So, basically, the second terms are important. The coefficients $1,2,5,12,...$ form the recurrence relation OEIS A000129:
$$a_1=1,a_2=2,a_n=2a_{n-1}+a_{n-2},$$
Its solution is:
$$a_n=\frac1{2\sqrt{2}}\left[(1+\sqrt{2})^n-(1-\sqrt{2})^n\right]$$
Hence:
$$a_n\sqrt{2}=\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2}=\frac{(1+\sqrt{2})^n+(1-\sqrt{2})^n-2(1-\sqrt{2})^n}{2}=\\
\frac{(1+\sqrt{2})^n+(1-\sqrt{2})^n}{2}-(1-\sqrt{2})^n$$
The first term is an integer, because its numerator is an even number and the second term approaches $0$ for large $n$, because $|1-\sqrt{2}|<1$, which were also stated in other answers and comments.
Hence, you can generalize for $\alpha=1+\sqrt{k}$:
$$a_n\sqrt{k}=\frac{(1+\sqrt{k})^n+(1-\sqrt{k})^n}{2}-(1-\sqrt{k})^n$$
So, it depends on how fast the second term $(1-\sqrt{k})^n$ approaches $0$.
