There are 101 terms in the first factor, of which 100 have a power of $x$ greater than 0. The second factor can be written:
$$
\sum_{k_1 + k_2 + k_3 = 101} 1^{k_1} \cdot x^{2 k_2} \cdot x^{k_3}
$$
(just imagine the product multiplied out) so you are asking how many sets of (k_1, k_2, k_3), all integers at least 0, are such that $k_1 + k_2 + k_3 = 101$ with $k_1 < 101$ (that one gives the term 1, there is just one of those).
Let's find out how many solutions $k_1 + k_2 + k_3 = 101$ has, this is like chopping a line of 101 $*$ into three pieces, say by separating with $|$ (this is called a stars and bars argument, for obvious reasons). But then you have a total of $101 + 2$ positions to be filled with 101 stars and 2 bars, that can be done in $\binom{101 + 2}{2} = 5253$ ways, of which you subtract 1 for $k_1 = 101$.
Combining your factors, you have 101 terms in the first factor, $5252$ with $x$ from the second, and the terms with $x$ in the product are the result of multiplying any term from the first factor with a term containing $x$ from the second, i.e., $101 \cdot 5252 = 53042$ (as long as no simpolifications happen).
A lower bound is that the result is a polynomial of degree $100 + 2 \cdot 101 = 302$, so there are at most $301$ terms with powers of $x$. But there are negative terms, so cancellation can/will happen, and you get less.
Hint. Note that $\sum_{k=1}^{20} k=210$ and
$$(x-1)(x^2 - 2)(x^3-3)\cdots(x^{20} - 20)=x^{210}
\left(1-\frac{1}{x}\right)
\left(1-\frac{2}{x^2}\right)
\left(1-\frac{3}{x^3}\right)\cdots\left(1-\frac{20}{x^{20}}\right).$$
Now consider the integer partitions with distinct parts of $210-203=7$:
$$7,\quad 6+1,\quad 5+2,\quad 4+3,\quad 4+2+1.$$
Best Answer
Try: $\begin{align*} 1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} \\ &= \frac{1 - x^{11}}{1- x} \\ (1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^4 \\ &= \frac{(1 - x^{11})^4}{(1- x)^4} \\ &= (1 - 4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44}) \cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\ [x^{10}] (1 - &4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44}) \cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\ &= [x^{10}] \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\ &= (-1)^{10} \binom{-4}{10} \\ &= \binom{10 + 4 - 1}{4 - 1} \\ &= 286 \end{align*}$