Let me indicate this for an infinite field of characteristic zero (just to avoid some separability problems). So, the claim is that if $I$ is an ideal in $k[x_1,\ldots, x_t]=R$ generated by polynomials of degree at most $d$ (I will use the usual notion of degree and if you want to use yours, multiply mine by $t$), and if $I$ is a proper ideal, then it is contained in a maximal ideal with the quotient field whose extension degree over $k$ is at most $d^t$, and this is optimal for general $k$.
As some of the commenters have pointed out, by taking general hyperplane sections, you may assume that $\dim R/I=0$ (and non-zero, which I will not repeat). Of course, $t$ may have become smaller, but does not affect our bound. Now, you can find a general linear combination of the generators of $I$, so that $t$ of them will define a scheme of dimension zero with ideal $J\subset I$. Let $J=(f_1,\ldots, f_t)$ and if $g_1,\ldots ,g_t$ are sufficiently general polynomials of degree at most $d$, then for a parameter $u$, one can assume that $J_u=(f_i+ug_i)$ define a family of finite schemes for `small' $u$, for all $u\neq 0$, $\dim_k R/J_u=d^t$, by Bezout's theorem. Now, pick an irreducible curve $C\subset \mathbb{A}^t\times D$, where $D$ is a suitable open set of the $u$-space which maps onto $D$. Then the map $C\to D$ is onto (if necessary making $D$ a bit smaller, but always not losing $u=0$) and for a general fiber $u=a$, the dimension of the fiber is bounded by $d^t$. But then, the special member also has the smae inequality.
In general, this is optimal. If there are elements $\alpha_i\in \overline{k}$ such that $[k(\alpha_i):k]=d$ and $[k(\alpha_1,\ldots,\alpha_t):k]=d^t$, then take $f_i(x_i)$ be the irreducible polynomial of $\alpha_i$. We see that $(f_1,\ldots, f_t)$ is a maximal ideal and the extension degree of the quotient is $d^t$, proving optimality.
