Sequence Problem

PROBLEM
Let
X₀ = 0,
X₁ = 1
and general formula for this sequence is
X_n+2 = 3·X_n+1 − 2·X_n
Prove that
Y_n = X_n² + 2⁽ⁿ⁺²⁾ are exact squares of some natural numbers.
PROOF
Let's prove by induction that X_n = 2ⁿ − 1
(a) For n=0 this is true since x₀ = 2⁰ − 1 = 0, as set above
(b) For n=1 this is true since x₁ = 2¹ − 1 = 1, as set above
(c) Assume that X_k = 2^k − 1 and X_k+1 = 2^k+1 − 1
and, based on that, calculate X_k+2:
X_k+2 = 3·X_k+1 − 2·X_k =
= 3·(2^k+1 − 1) − 2·(2^k − 1) =
= 3·2^k+1 − 3 − 2·2^k + 2 =
= 3·2^k+1 − 2^k+1 − 1 =
= 2·2^k+1 − 1 =
= 2^k+2 − 1
which corresponds to a formula that we are trying to prove.

Using this general formula for X_k = 2^k − 1, we will prove that
Y_n = X_n² + 2⁽ⁿ⁺²⁾ are exact squares of some natural numbers.
Indeed,
Y_n = (2ⁿ − 1)² + 2⁽ⁿ⁺²⁾ =
= 2²ⁿ − 2·2ⁿ + 1 + 2⁽ⁿ⁺²⁾ =
= 2²ⁿ − 2·2ⁿ + 1 + 4·2ⁿ =
= 2²ⁿ + 2·2ⁿ + 1 =
= (2ⁿ + 1)²