For $k=1$, observe that there is only one solution of $a^x-b^y=1$, namely $3^2-2^3=1$.
This is Catalan's conjecture, proved by Mihailescu.

For $k\geq 3$, there is nice paper
"Small gaps between almost primes, the parity problem and some
conjectures of Erdos on consecutive integers" by
D. A. GOLDSTON, S. W. GRAHAM, J. PINTZ AND C. Y. YILDIRIM.
International Mathematics Research Notices, Vol. 2011, No. 7, pp. 1439–1450
http://www.math.boun.edu.tr/instructors/yildirim/paper/GGPY3imrn.pdf

with many interesting results:

Theorem 1.1. Let $A$ be any multiset of positive integers that contains $\{2, 1, 1, 1\}$ as a
subset. There exist infinitely many integers $x$ such that $x$ and $x + 1$ both have exponent
pattern $A$. Consequently, for any integer $B \geq 0$, there exist infinitely many integers $x$
such that
$\omega(x) = \omega(x + 1) = 4 + B, \Omega(x) = \Omega(x + 1) = 5 + B$, and $d(x) = d(x + 1) = 24 \, 2^B$.
...

Theorem 1.2. There are infinitely many integers $x$ with
$\omega(x) = \omega(x + 1) = 3$.


Theorem 1.3. There are infinitely many integers x with
$\Omega(x) = \Omega(x + 1) = 4$.

And many similar results.

From this it seems that only the case $k=2$ of your question is left. You give  one solution, which settles your question.

The more difficult question if there are infinitely many solutions with $k=2$ may be open. But let me mention a partial result:

D. A. Goldston, S. W. Graham, J. Pintz and C. Y. Yildirim
Small gaps between primes or almost primes
Journal: Trans. Amer. Math. Soc. 361 (2009), 5285-5330
https://www.ams.org/journals/tran/2009-361-10/S0002-9947-09-04788-6/home.html

Theorem 3. Let $q_n$ denote the $n$-th number that is a product of exactly two primes.
Then $\lim \inf_{n \rightarrow \infty} (q_{n+1} - q_n) \leq 26$.