We shall use the following assertion.

Lemma. Let $k \ge 2$ be a positive integer. Then the equation $2x^{2k} + 1 = y^2$ does not have solutions in positive integers.

Proof. Assume that $x, y$ and $k \ge 2$ are positive integers such that $2x^{2k} + 1 = y^2$ and $x$ is minimum possible. It is obvious that $x$ is even and $y$ is odd. Let us denote $x = 2a$ and $y = 2b + 1$. Then $2^{2k-1}a^{2k} = b(b+1)$ and $(b, b+1) = 1$. There are two possibilities:
- if $b = x_1^{2k}$ and $b+1 = 2^{2k-1}x_2^{2k}$, $x_1, x_2 \in \mathbb{N}$, $x_1x_2 = a$, then $2^{2k-1}x_2^{2k} - x_1^{2k} = 1$, which gives a contradiction modulo 4;
- if $b = 2^{2k-1}x_1^{2k}$ and $b+1 = x_2^{2k}$, $x_1, x_2 \in \mathbb{N}$, $x_1x_2 = a$, then $x_2^{2k} - 2^{2k-1}x_1^{2k} = 1$, which leads to the equation $y_1^2 = 2^{2k-1}x_1^{2k} + 1$, $y_1 = x_2^k$, where

we notice that $x_1 < x$.
It is clear that the above argument of decreasing the degrees of 2 can be continued until we have degree at most 5. Therefore we reach the equation $y_0^2 = 8x_0^{2k} + 1$, where $x_0 < x$ and $y_0 = y_2^k$, $y_2 \in \mathbb{N}$. Clearly, $y_0$ is odd and we set $y_0 = 2c+1$. We obtain $c(c+1) = 2x_0^{2k}$, where $(c, c+1) = 1$. We have again two possibilities:
- if $c = x_3^{2k}$ and $c+1 = 2x_4^{2k}$, $x_3, x_4 \in \mathbb{N}$, $x_3x_4 = x_0$, then $4x_4^{2k} = 2c+2 = y_2^k+1$, whence $(2x_4^k-1)(2x_4^k+1) = y_2^k$. This leads to $2x_4^k-1 = y_3^k$, $2x_4^k+1 = y_4^k$, $y_3, y_4 \in \mathbb{N}$, $y_3y_4 = y_2$, and finally $y_4^k - y_3^k = 2$, which is impossible;
- if $c = 2x_3^{2k}$ and $c+1 = x_4^{2k}$, $x_3, x_4 \in \mathbb{N}$, $x_3x_4 = x_0$, then $2x_3^{2k}+1 = (x_4^k)^2$, which contradicts to the choice of $x$ as minimal.
This completes the proof of the lemma.

The roots of the characteristic equation $t^2 - 6t + 1 = 0$ of our sequence are $t_{1,2} = 3 \pm 2\sqrt{2}$. Therefore we find (using the conditions $a_1 = 2$ and $a_2 = 12$)
$$a_n = \frac{(3 + 2\sqrt{2})^n - (3 - 2\sqrt{2})^n}{2\sqrt{2}}.$$
Denote $(3 + 2\sqrt{2})^n = \alpha_n + \beta_n\sqrt{2}$, $\alpha_n, \beta_n \in \mathbb{N}$. Then $(3 - 2\sqrt{2})^n = \alpha_n - \beta_n\sqrt{2}$, $\alpha_n = \beta_n$ and $\alpha_n^2 - 2\beta_n^2 = 1$. Now, if $a_n$ is perfect power for some $n$, then the last two equalities give a contradiction with the lemma.