We can exactly express the count $p(n)$ of $n$-digit numbers from $\{1,2,3,4,5\}$ where adjacent digits cannot differ by less than two, as WimC outlines above. If $p_k(n)$ denotes the count of such numbers with leading digit $k$, then for $n \ge 1$:
$p(n) = p_1(n) + p_2(n) + p_3(n) + p_4(n) + p_5(n)$
Furthermore since an $n+1$ digit number of the required type consists of an $n$ digit such number with a compatible new leading digit (one that differs by at least two from the previous digit), we have by induction the matrix expression:
$p(n) = u A^{n-1} u'$
where row $u = \begin{pmatrix} 1&1&1&1&1 \end{pmatrix}$, and $A$ is WimC's $5\times5$ Toeplitz matrix. For example, $p(1) = 5$, $p(2) = 12$, $p(3) = 30$, $p(4) = 74$, and $p(5) = 184$.
The desired bounds can be framed in terms of Rayleigh quotients $R_n = \frac{u A^n u'}{u u'}$:
$2.4^n \le R_n \le 2.5^n$
because the denominator $u u' = 5$, so that $p(n) = 5 R_{n-1}$. The upper bound is implied by the computation of the dominant eigenvalue of real symmetric $A^n$, which is the $n$th power of the dominant eigenvalue of $A$.
As WimC noted, $A$ has a dominant eigenvalue $\lambda_{max} \approx 2.4812 \lt 2.5$. Indeed $A$'s characteristic polynomial is $(\lambda^3 -2\lambda^2 -2\lambda + 2)(\lambda + 2)\lambda$ and there are five distinct real eigenvalues. By virtue of $A$ being real symmetric, these correspond to mutually orthogonal eigenvectors.
The two "easy" eigenvalues $\lambda = -2,0$ correspond to unit (left) eigenvectors $\frac{1}{2}\begin{pmatrix} 1&1&0&-1&-1 \end{pmatrix}$ and $\frac{1}{2}\begin{pmatrix} 1&-1&0&1&-1 \end{pmatrix}$ respectively. These happen to be orthogonal to $u$, and so are not used below to express $u$ in terms of the orthonormal basis of $A$-eigenvectors.
$ \lambda_{max} \approx 2.4812, v_{max} \approx \begin{pmatrix} 0.52990 & 0.35775 & 0.42713 & 0.35775 & 0.52990 \end{pmatrix}$
$ \lambda_{med} \approx 0.68889, v_{med} \approx \begin{pmatrix} 0.17934 & -0.57645 & 0.52066 & -0.57645 & 0.17934 \end{pmatrix}$
$ \lambda_{min} \approx -1.17009, v_{min} \approx \begin{pmatrix} 0.43249 & -0.19929 & -0.73924 & -0.19929 & 0.43249 \end{pmatrix}$
Taking the dot-product of $u$ with each of these vectors gives us the coefficients of the basis expansion:
$ u \approx 2.20243 v_{max} - 0.27356 v_{mid} - 0.27284 v_{min} $
Thus the Rayleigh quotient $R_n$ can be computed from that expansion:
$ A^n u' \approx 2.20243 \lambda_{max}^n v_{max} - 0.27356 \lambda_{med}^n v_{mid} - 0.27284 \lambda_{min}^n v_{min} $
$ R_n \approx (2.20243^2 \lambda_{max}^n + 0.27356^2 \lambda_{med}^n + 0.27284^2 \lambda_{min}^n)/5 $
The lower bound $R_n \ge 2.4^n$ may be established by noting that it holds with equality for $n=0,1$ and then showing that $2.4^{-n} R_n$ increases thereafter:
$ 2.4^{-n} R_n \approx 0.97014 (\frac{\lambda_{max}}{2.4})^n + 0.01497 (\frac{\lambda_{med}}{2.4})^n + 0.01489 (\frac{\lambda_{min}}{2.4})^n $
The growing leading term, whose base $\frac{\lambda_{max}}{2.4}$ exceeds 1, will of course eventually dominant the other two terms, one of which is positive but shrinking and the other alternating in sign (but also shrinking).
Treating this as a continuous function of the exponent n (over positive reals) and differentiating gives a concrete proof that it increases for $n \gt 1$, which agrees with the several initial values already tabulated:
$2.4^{-1} R_1 = 1$
$2.4^{-2} R_2 = 1.041666...$
$2.4^{-3} R_3 = 1.07060185...$
$2.4^{-4} R_4 = 1.109182098765432...$