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Given a multivariate polynomial $p(x_1,\ldots,x_n)$ with real coefficients, its real zero set, i.e. the set of real roots of the polynomial, is $\{(x_1,\ldots,x_n) \in \mathbb{R}^n \mid p(x_1,\ldots,x_n) = 0\}$.

What characterizations (necessary and sufficient conditions) are known for this set to be bounded?

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Let us call the solution set to be $V$. We can consider this set in $\mathbb{RP}^n$ where $V \subset \lbrace x_0 \neq 0 \rbrace$. Let us call the homogenized polynomial $\tilde{p}$. Then $V(\tilde{p}) = \overline{V}$ in Zariski topology.

If the set $V$ is unbounded, then the closure of $V$ in real topology has non-empty intersection with hyperplane at infinity that is the set $\lbrace x_0 = 0 \rbrace$. This implies that $\overline{V} \cap \lbrace x_0 = 0 \rbrace \neq \emptyset$.

Assume that $V$ is irreducible and $\overline{V} \cap \lbrace x_0 = 0 \rbrace \neq \emptyset$, then we get a sequence of points $p_i V$ such that $p_i$ converges to $p$ in hyperplane at infinity. Let $p_i = [p_{0i} : p_{1i} : \dots : p_{ni}]$, then one of the ratios $p_{ji}/p_{0i}$ is unbounded(since the point $[0 : 0 : \dots : 0]$ is not a point). Thus we get that the solution set is unbounded. Now if $V$ is not irreducible, then we can argue similarily for each of the irreducible components.

Thus the $V$ is bounded iff $\overline{V} \cap \lbrace x_0 = 0 \rbrace = \emptyset$. This holds iff the highest degree homogeneous term of the polynomial $p$ has no zeros except $(0, \dots , 0)$.

For example : Let $p(X_1,X_2) = X_1^2 + X_2^2 - 1$. Then $\tilde{p}(X_0,X_1,X_2) = X_1^2+X_2^2 - X_0^2$. The zeros of this polynomial with $X_0 = 0$ is $(0,0)$ and hence $V(p)$ must be compact, which clearly is true.

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