Alternative to binomial expansion, we can deduce that $\rm\:w = (9+4\sqrt{5})^n + (9-4\sqrt{5})^n\:$ is even by noting that the notion of *parity* uniquely extends from $\;\mathbb Z\:$ to $\:\mathbb Z[\sqrt{5}]\:$ by defining $\rm\:\sqrt{5}\:$ to be odd. Conjugating shows $\rm\:w'= w,\:$ therefore $\rm\:w\in \mathbb Z\,$ with parity $\rm\ {odd}^n\! + {odd}^n = odd+odd = even.\ $ The rest follows easily from $\,0<9-4\sqrt{5}<1\,$ as Beni explained. Notice, in particular, how this viewpoint is a very *natural* higher-degree extension of ubiquitous parity-based proofs in $\mathbb Z.$

Alternatively, it follows immediately from the fact that, mod $2$, the sequence $\rm\:w_n\:$ satisfies a monic integer-coefficient recurrence of order $\,\rm\color{#c00}2\,$ and the first $\,\rm\color{#c00}2\,$ terms are $\equiv 0\pmod{2}.\:$ Hence by induction so are all subsequent terms, i.e. $\rm\ f_{n+2} \equiv\ a\ \color{#0a0}{f_{n+1}} + b\,\color{#0a0}{f_{n}} \equiv\ 0\ $ since by induction $\rm\ \color{#0a0}{f_{n+1},\ f_{n}} \equiv 0.\: $ Or, said equivalently $\rm\:f_n \equiv 0\ $ by the *uniqueness theorem* for solutions of difference equations (recurrences). As I frequently emphasize uniqueness theorems provide very powerful tools for proving equalities.

Worth emphasis: for difference (vs. differential) equations, the uniqueness theorem is trivial, amounting to an obvious induction that if two solutions of a order $\rm\:d\:$ monic integer coefficient recurrence agree for $\rm\:d\:$ initial values then they agree at all subsequent values; equivalently, taking differences, if a solution is $\:0\:$ for $\rm\:d\:$ initial values then it is identically $\,0.\,$ The induction step simply employs the recurrence to lift up the equality from the prior $\rm\,d\,$ values - as above for $\rm\,d=2.$

More generally the same holds true for integer-linear combinations of roots of any monic integer coefficient equation (i.e. algebraic integer roots) since they too will satisfy a monic integer coefficient recurrence, viz. the *characteristic equation* associated to the polynomial having said roots (the quadratic case is the widely studied Lucas sequence). Thus every term of the sequence will be divisible by $\rm\:m\:$ iff it is true for the first $\rm\:d\:$ (= order) terms. More generally one easily checks that the gcd of all terms is simply the gcd of the initial values.

Note that, as above, one requires only the knowledge of the *existence* of such a recurrence. There is no need to explicitly calculate the coefficients of the recurrence; rather, only its order is employed.