I have the following constructive NT-flavored solution, but I'm not 100% certain about it. Please point out any possible inaccuracies! Firstly, some better notation. We write $a_n^M=\Delta^Ma_n$, and we have the relation $a_n^{M}+a_n^{M+1}=a_{n+1}^M$. We can think of this as "gluing together" Pascal triangles in the plane, which induces us to try and find some explicit formula based on binomials. I claim $a_n^M = \sum_{k \in \mathbb Z} \binom{n}{kt-M}$ will satisfy what we want, where $t=2022$ (the generalized binomial coefficiente $\binom{x}{y}, x \geq 0, y \in \mathbb Z$ is said to be zero whenever $y < 0$ or $y > x$. In particular, the infinite sum is only nonzero for $k$ in a bounded interval). Stifel, i.e., $\binom{x}{y-1}+\binom{x}{y}=\binom{x+1}{y}$ still holds, and so it is imediate to verify $a_n^M$ satisfies the patriota relation, as $\sum_{k \in \mathbb Z} \binom{n}{kt-M} + \sum_{k \in \mathbb Z} \binom{n}{kt-M-1} = \sum_{k \in \mathbb Z} \binom{n+1}{kt-M}$. Moreover, for $k=l=0$ we have $a_n^t=\sum_{k \in \mathbb Z} \binom{n}{kt-t} = \sum_{k \in \mathbb Z} \binom{n}{k(t-1)} = \sum_{k \in \mathbb Z} \binom{n}{kt} = \sum_{k \in \mathbb Z} \binom{n}{kt-0} = a_n^0$, so the sequence indeed is patriota for $t$. Now, suppose there exist $p,q,s \in \mathbb N, 0 < s < t$ such that $a_{n+p}=a_{n+q}^s$. Computing directly in our formula, we get that $\sum_{k \in \mathbb Z} \binom{n+p}{kt}=\sum_{k \in \mathbb Z} \binom{n+q}{kt-s}$, and if we change $n$ by $n+1$ and calculate the difference of these two relations, we get by Stifel $\sum_{k \in \mathbb Z} \binom{n+p}{kt-1}=\sum_{k \in \mathbb Z} \binom{n+q}{kt-s-1}$. In general, we can write $\sum_{k \in \mathbb Z} \binom{n+p}{kt-m}=\sum_{k \in \mathbb Z} \binom{n+q}{kt-s-m}$ for each $m \in \{0,1,\dots, t-1\}$. Summing over all this $m$, we imediately get as we run through all residues $\bmod\,t$, that $\sum_{j \in \mathbb Z} \binom{n+p}{j}=\sum_{j \in \mathbb Z} \binom{n+q}{j} \iff 2^{n+p}=2^{n+q} \iff p=q$. We have $\sum_{k \in \mathbb Z} \binom{n}{kt}=\sum_{k \in \mathbb Z} \binom{n}{kt-s}$ for all big $n$. Recall that $k=6r$, for $r=337$ a prime and choose $n=6r^l$, for big enoguh $l>1$. We analyse this last relation $\pmod r$. By Lucas' Theorem, we have a cute lemma: $\nu_r(\alpha) > \nu_r(\beta) \implies \binom{\alpha}{\beta} \equiv 0 \mod r$ if $\alpha, \beta > 0$. The only binomials candidates not to vanish in LHS are the ones for which $\nu_r(6rk) \geq \nu_r(6r^l) \iff \nu_r(k) \geq l-1$ or $k=0$. But we also have that out of bounds $k$ will lead to generalized binomial coefficients vanishing directly, so the additional condition $0 \leq 6rk \leq 6r^l \iff 0 \leq k \leq r^{l-1}$ must be satisfied. So, \[\sum_{k \in \mathbb Z} \binom{n}{kt} \equiv \sum_{k \in \{0, r^{l-1}\}} \binom{6r^l}{6rk} \equiv 1 + 1 \equiv 2 \pmod p\] Now we look at the RHS. $0$ is never attained in the down part of the binomial, because $t \nmid s$. So a non-vanishing binomial coefficient satisfies $\nu_r(6rk-s) \geq \nu_r(6r^l) \implies 6k-s'=xr^{l-1}$, for some $x$, where we used $s'=s/r$. As $r \equiv 1 \pmod 6$, it follows $x+s' \equiv 0 \pmod 6$. By $0 < xr^l < 6r^l$, $0 < x < 6$. By $0 < s < 6r$, $0 < s' < 6$. From these we get $x+s'=6$, therefore if there is such an $x$ it is unique and belongs to $\{1,2,3,4,5\}$. But again by Lucas' Theorem, $\binom{6r^l}{xr^l} = \binom{6}{x}\binom{0}{0}^l = \binom{6}{x} \in \{6, 15, 20\}$ and none of these is $2 \pmod r$.

Congratulations!

One slick solution (found by my students) is first considering the sequence $b_{n+M} - b_n = b_{n+1}$, with $b_0,b_1,\ldots,b_{M-1}$ large consecutive numbers (say, $M,M+1,\ldots,2M-1$ respectively). Then $b_n$ is increasing, as $b_M = b_0+b_1 = 2M+1 > b_{M-1} = 2M-1$ and, for $n>M$, $b_n = b_{n-M} + b_{n-M+1} > b_{n-M-1} + b_{n-M} = b_{n-1}$. In particular, all numbers in the sequence $b_n$ are distinct. Then $a_n = b_{nM}$. One can prove that $\Delta^i a_n = b_{i+nM}$ (induct on $i$), so $a_n,\Delta^1 a_n,\ldots,\Delta^{M-1}a_n$ partition $b_n$. Since all terms in $b_n$ are distinct, $\Delta^ia_n$ does not work. However, $\Delta^M a_n = b_{M+nM} = a_{n+1}$ works and we're done.

Thanks..Nice...

(I mistunderstood whoops)