For any set $S$ let $\mu_x(S)=\sum_{s\in S} s^x$. When $S$ is a nonempty subset of $\lbrace 1,\ldots,k\rbrace$, $1\le \mu_x(S)\le k^{x+1}$, so the number of possible values of $(\mu_1(S),\mu_2(S),\ldots,\mu_t(S))$ is at most $k^2\times\cdots\times k^{t+1}= k^{t(t+3)/2}$.

Since there are $2^k-1$ non-empty subsets of $\lbrace 1,\ldots,k\rbrace$, some pigeons tell us that when $2^k-1 > k^{t(t+3)/2}$ there are two different nonempty subsets $S,S'$ of $\lbrace 1,\ldots,k\rbrace$ such that $(\mu_1(S),\mu_2(S),\ldots,\mu_t(S))=(\mu_1(S'),\mu_2(S'),\ldots,\mu_t(S'))$. By deleting common elements of $S$ and $S'$ we can assume they are disjoint and have at most $k$ elements altogether.

Now $\sum_{s\in S} s^x=\sum_{s\in S'} s^x$ implies $\sum_{s\in S} (s/k!)^x=\sum_{s\in S'} (s/k!)^x$ and $k!/s$ is an integer for $s\in \lbrace 1,\ldots,k\rbrace$.
Therefore, if $2^k-1 > k^{t(t+3)/2}$, there are disjoint integer sequences $(a_1,\ldots,a_m), (b_1,\ldots,b_n)$ with $m+n\le k$ such that $x=1,2,\ldots,t$ are all solutions.

For any $t$, the inequality $2^k-1 > k^{t(t+3)/2}$ holds for large enough $k$, approximately $k\gt t^2\log_2 t$.

This argument can be tightened in various ways. Solutions with $m=n$ need only slightly larger $k$.

To get sharper bounds, note that for random subsets $S$, $(\mu_1(S),\mu_2(S),\ldots,\mu_t(S))$ concentrates near its ($t$-dimensional) mean. This implies that some box gets two pigeons rather sooner than the simple calculation gave.