By the triangle inequality, $a < b + c$, hence $a < 2b$, using $b = c$. Therefore $a^2 < 4bc$, as required.

If $x > 1$, and $a = x^2 + 1$, $b = 2x$, $c = x^2 - 1$, then
$$b^2 + c^2 = 4x^2 + (x^4 - 2x^2 + 1) = x^4 + 2x^2 + 1 = a^2,$$
and so $a$, $b$, $c$ are side lengths of a right-angled triangle with hypotenuse $a$. Then
$$\begin{aligned} a^2 - 4bc &= (x^2 + 1)^2 - 8x(x^2 - 1) \\ &= x^4 - 8x^3 + 2x^2 + 8x + 1 \\ &= x^3(x - 8) + 2x^2 + 8x + 1. \end{aligned}$$
Clearly, the polynomial $x^3(x-8)+2x^2+8x+1$ is positive if $x$ is large enough, e.g., any $x \ge 8$ will do. So, as long as $x \ge 8$ we have a right-angled triangle in which $a^2 > 4bc$.

Alternatively, we may find such examples as follows. If $a$ is the length of the hypotenuse of the right-angled triangle $ABC$, its area ($ABC$) is equal to $bc/2$. Hence, the desired inequality $a^2 > 4bc$ is equivalent to $a^2 > 8(ABC)$. If $h$ is the height of triangle $ABC$ on the hypotenuse $a$, then ($ABC$) = $ah/2$ and the desired inequality becomes $a > 4h$. In the diagram below, this means that $A$ should be on the semicircle but not above the dashed line.

To make this more explicit, let the foot of the altitude from $A$ divide the hypotenuse into segments of length $p$ and $q$, and recall that $h^2 = pq = p(a-p)$. Thus, we want $a^2 > 16h^2 = 16p(a-p)$, i.e. $a^2 - 16pa + 16p^2 > 0$. This is equivalent to
$$\left(\frac{a}{p} - 8\right)^2 > 48, \text{ i.e. } \left|\frac{a}{p} - 8\right| > 4\sqrt{3}.$$
As $0 < p < a$ in our case, given $a$ we can choose any $p$ such that
$$0 < p < \frac{2 - \sqrt{3}}{4} \quad \text{or} \quad \frac{2 + \sqrt{3}}{4} < p < a.$$
Because $c^2 = p^2 + h^2 = p(p+q) = ap$ and $b^2 = aq = a(a-p)$, the value of $p$ determines $b = \sqrt{a^2 - ap}$ and $c = \sqrt{ap}$ for any given $a$. For example, with $a = 41$ and $p = 81/41$, we obtain $b = 40$ and $c = 9$. In this example, $a^2 = 1681 > 1440 = 4bc$.