Let's start with $x \in \mathbb{R}^d$, $y \in \mathbb{R}^m$, $H \in \mathbb{R}^{m,d}$ and consider
$$F:x \mapsto \|Hx - y\|_2^2 $$
I want to examine whether or not $F$ is coercive, i.e. if for all sequences $(x^n)_{n \in \mathbb{N}} \subset \mathbb{R}^d$ with $\lim_{n \to \infty}\|x^n\|_2 \to +\infty$ $\implies$ $\lim_{n \to \infty}F(x^n) \to +\infty$. We can see for instance very easily that, if we choose $y=0$ and $H=0$ or in fact any column of $H$ equal to zero, coercivity does not hold.
This observation leads to the following claim: If $H$ is invertible, then $F$ is coercive. Is there an easy way to see why this holds?
Indeed, this is true. In fact, you only need $H^T H$ invertible (equivalently, $\operatorname{rank} H = d$).
The most common way to establish coercivity is to find a lower bound for $F(x)$ in terms of $\|x\|$.
In our case, expand $$ F(x) = x^T H^T H x - 2y^T Hx + \|y\|_2^2. $$ You can complete the square and write $$ F(x) = (x - \xi)^T H^T H (x - \xi) + \text{constant}, $$ where $\xi = (H^T H)^{-1} H^T y$. Observe that $H^T H$ is positive semidefinite. And since it is invertible, it's actually positive definite. So all eigenvalues are strictly positive. Let $\lambda_d$ be the smallest eigenvalue of $H^T H$. Then $$ F(x) \geq \lambda_d \|x - \xi\|_2^2 + \text{constant}. $$ This lower bound establishes coercivity.