Growth of a linear recurrent sequence

First note that $\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\in\mathbb{Q}(\sqrt{-2})$ is not a root of unity, because it does not equal $\pm 1$. Therefore Baker's famous theorem shows that, for some effectively computable constant $c>0$, $$\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^m-1\right|>m^{-c},\qquad m\geq 2.\tag{$*$}$$ (The variables $m$ and $n$ are integers in this post.) Applying this for $m=2n$, we get $$\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^n-1\right|\cdot\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^n+1\right|>(2n)^{-c},\qquad n\geq 1.$$ The first factor is less than $2$, hence multiplying both sides by the absolute value of $(1-i\sqrt{2})^n$, we infer $$\Bigl|(1+i\sqrt{2})^n+(1-i\sqrt{2})^n\Bigr|>\frac{1}{2}\cdot\frac{3^{n/2}}{(2n)^c},\qquad n\geq 1.$$ This means that the sequence $(a_n)$ grows exponentially: $$ |a_n|>\frac{1}{4}\cdot\frac{3^{n/2}}{(2n)^c},\qquad n\geq 1.$$ In particular, there are only finitely many $n$'s with $|a_n|\leq 100$, and these can be effectively bounded. Then, up to that bound, one can check with a computer (at least in theory) which $n$'s satisfy $|a_n|\leq 100$.

Added. In ($*$), the exponent $c=5\times 10^9$ is admissible by Corollary 2.3 in Matveev: An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers. II (Izv. Math. 64 (2000), 1217-1269). This means that $|a_n|\leq 100$ implies $n<3\times 10^{11}$.

Besides the archimedean technique from Baker's theorem in another answer, this type of question can be handled both qualitatively and quantitatively using $p$-adic methods. This was illustrated, for the sequence you asked about, on https://math.stackexchange.com/questions/873147/finding-non-negative-integers-m-such-that-1-sqrt-2m-has-real-part/873529 with the task being a determination of when $a_n = \pm 1$. I wrote up an account of this in another $p$-adic field at http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/strassmannapplication.pdf.