Austria competition problems

Answer. Yes.

Sequence $(a_n)$ starts with values $2$, $14$, $198$, $2786$, $39202$, $551614$. Sequence $(b_n)$ starts with values $2$, $14$, $82$, $478$, $2786$, $16238$, $94642$, $551614$. We therefore conjecture that $a_{2k+1} = b_{3k+1}$ holds for $k \ge 0$.
Shifting the recurrence yields
$$a_{n+2} - 14a_{n+1} - a_n = 0, \\ a_{n+1} - 14a_n - a_{n-1} = 0, \\ a_n - 14a_{n-1} - a_{n-2} = 0$$
for $n \ge 2$. Multiplying these recurrences by $1$, $14$ and $-1$, respectively, and taking the sum yields $a_{n+2} - 198a_n + a_{n-2} = 0$ and thus
$$a_{n+2} = 198a_n - a_{n-2}$$
for $n \ge 2$.

Shifting the recurrence of $(b_n)$ yields
$$\begin{align*} b_{n+3} - 6b_{n+2} + b_{n+1} &= 0, \\ b_{n+2} - 6b_{n+1} + b_n &= 0, \\ b_{n+1} - 6b_n + b_{n-1} &= 0, \\ b_n - 6b_{n-1} + b_{n-2} &= 0, \\ b_{n-1} - 6b_{n-2} + b_{n-3} &= 0 \end{align*}$$
for $n \ge 3$. Multiplying these recurrences by $1$, $6$, $35$, $6$ and $1$, respectively, and taking the sum yields $b_{n+3} - 198b_n + b_{n-3} = 0$ and thus
$$b_{n+3} = 198b_n - b_{n-3}$$
for $n \ge 3$.
We see that the subsequences $(a_{2k+1})$ and $(b_{3k+1})$ have the same initial values $a_1 = b_1 = 14$ and $a_3 = b_4 = 2786$ and fulfil the same recurrence. This implies that $a_{2k+1} = b_{3k+1}$ for all $k \ge 0$.
From the given recurrence, it is obvious that the sequence $(a_n)$ is strictly increasing. Thus we also get infinitely many values which occur in both sequences.