We are provided a single Diophantine equation $$f(x_1,\dots,x_n)=0$$ having degree $\geq2$ and having the promise it has $\leq1$ solutions in the set $\{0,\dots,m-1\}^n$ and $t$ is the number of terms in the polynomial.

We are to decide if $$|\{(x_1,\dots,x_n)\in\{0,1\}^n:f(x_1,\dots,x_n)=0\}|>0?$$

Is there a $\mathsf{poly}(mnt)$ algorithm for the problem?

If $m=2$ degree of every $x_1$ to $x_n$ can be reduced to $1$.

I think for general $m$ it should be in $\mathsf{poly}(mnt)$.

Valiant-Vazirani is applicable but unless $t=\mathsf{poly}(n)$ a $\mathsf{poly}(nt)$ algorithm cannot resolve $\mathsf{NP}$ versus $\mathsf{BPP}$. It is not clear $\mathsf{SAT}$ is Valiant-Vazirani reducible to a $\mathsf{PromiseDiophantine}$ problem of having $\leq1$ integral solutions.