Find the values of $k$ for which the curve $y = 2x^2 + (3k + 1)x + 4$ lies entirely above the line $y = 2x - 3k^2 + 5$.

1. Form the inequality for the curve to lie entirely above the line and rearrange it into standard quadratic form.

For the curve to lie entirely above the line, we must have:
$2x^2 + (3k + 1)x + 4 > 2x - 3k^2 + 5$
$2x^2 + (3k + 1 - 2)x + 4 - 5 + 3k^2 > 0$
$2x^2 + (3k - 1)x + 3k^2 - 1 > 0$

2. State the condition for the quadratic expression to be always positive.

Since the coefficient of $x^2$ is $2 > 0$, the quadratic expression is always positive for all real $x$ if and only if its discriminant is negative:
$b^2 - 4ac < 0$

3. Substitute the coefficients into the discriminant and simplify.

$(3k - 1)^2 - 4(2)(3k^2 - 1) < 0$
$9k^2 - 6k + 1 - 8(3k^2 - 1) < 0$
$9k^2 - 6k + 1 - 24k^2 + 8 < 0$
$-15k^2 - 6k + 9 < 0$

4. Solve the resulting quadratic inequality for $k$.

Divide the inequality by $-3$ (and flip the inequality sign):
$5k^2 + 2k - 3 > 0$
Factorise the quadratic:
$(5k - 3)(k + 1) > 0$
The critical values are $k = \frac{3}{5}$ and $k = -1$.
Since the inequality is $> 0$, the solution is:
$k < -1 \text{ or } k > \frac{3}{5}$