Are the sets $\mathbb X=\{0,1,4,15,56,...,x_h,...\} $ and $\mathbb Y=\{0,2,12,70,408,...,y_i,...\}$ (excepting the elements $x_0=y_0=0$) disjunct? In trying to answer another question I came to the problem as written in the title:    

Q1: Are the sets $\mathbb X=\{0,1,4,15,56,...,x_h,...\} $ and $\mathbb Y=\{0,2,12,70,408,...,y_i,...\}$ (excepting the elements $x_0=y_0=0$) disjunct?     

and how to approach a proof.      
I have two ways to describe the sets as sequences with their elements depending on their indexes: 
 $$x_0=0, x_1=1, x_{h+1}=4x_h-1x_{h-1}\\
 y_0=0, y_1=2, y_{i+1}=6y_i-1y_{i-1}\\ \tag 1$$
and using $p=2+\sqrt3$ and $q=3+\sqrt8$
 $$ x_h = f(h)= \sinh (h \cdot \ln p)/\sqrt 3 \\
 y_i =g(i) = \sinh (i \cdot \ln q)/\sqrt 8  \tag 2$$
(The latter is a compacted version of a Binet-like formula (as it is known for instance for the Fibonacci numbers))        
What I've tried was ("ansatz 1") to look at the composition of $i=g°^{-1}(f(h))$ and see, whether I can prove, that $i$ is never integer when $h$ is integer, but looking at heuristics and working with the results a bit I do not see any further forcing and/or reliable way to proceed.        
Next ("ansatz 2") I've looked at the prime-factorizations of $f(h)$ and $g(i)$ in terms of $h$ resp. $i$. This gives at least insight to some basic facts, for instance that $h$ must be even, then divisible by $3$ and from this $i$ must be divisible by $3$ and so on, and at least leading to a proof for small $h$ and $i$. But this is so far only reasoning for finitely many classes of cases.         

Q2: Can my ansatz 1 or ansatz 2 be improved? Or not?
  Q3: Is there any differnt route to approach this?

 A: A simple observation seems to contain the potential of a more immediate proof than that provided in the answer of @RenéGy.

*

*The set $X$ is the set of solutions of the Pell-equation (taken from the referred question):
$$ 3x^2+1 = a^2 \tag {1.1}$$
This gives the set of solutions, depending of index $h \in \{0,1,2,3,..\}$
$$ x_h = \{0,1,4,15,...\}_h \\ 
   a_h = \{1,2,7,26,...\}_h\\ \tag {1.2}
$$
both with the same recursion $x_{h+1}=4x_h-x_{h-1}$ resp. $a_{h+1}=4a_h-a_{h-1}$


*The set $Y$ is the set of solutions of the Pell-equation:
$$ 2y^2+1 = b^2 \tag {2.1}$$
This gives the set of solutions, depending of index $i \in \{0,1,2,3,..\}$
$$ y_i = \{0,2,12,70,...\}_i \\ 
   b_i = \{1,3,17,99,...\}_i\\\tag {2.2}
$$
both with the same recursion $y_{i+1}=6y_i-y_{i-1}$ resp. $b_{i+1}=6b_i-b_{i-1}$

If we assume some element $x_h$ exists as well as element $y_i$ then we could equate
$$ 3x_h^2 + 1=  a_h^2 \qquad \text{  and at the same time  } \qquad  2x_h^2 + 1= b_i^2 \\
\implies x_h^2 = a_h^2-b_i^2 \tag 3$$
This is in the form of a pythagorean triple, and solutions for that must have the form
(rewritten from the formula in wikipedia)
$$ x_h=(m^2-n^2) \qquad b_i=2mn \qquad a_h = (m^2+n^2) \qquad \text{ with } m \gt n \gt 0 \tag 4
$$
This formula shows, that the parity of $x_h$ and $a_h$ must be of the same type since $m^2-n^2$ and $m^2+n^2$ have the same parity.
Indication of nonexistence: From the table in $(1.2)$ we see, that $x_h$ and $a_h$ for $h \gt 0$ have always opposite parity, contradicting $(4)$. Thus no solution $x_h = y_i$ for $h,i \gt 0$ should exist.

So if I didn't mess something we need only the proof of the opposite parity of $x_h$ and $a_h$ and this seems really simple because of the definitions by the recursion-formula.
