When are the eigenvalues of a matrix containing all squared elements irrational/rational?

The claim is not true. The matrix $$\begin{bmatrix}1^2&36^2\\5^2&26^2\end{bmatrix}$$ has eigenvalues $721$ and $-44$, which are evidently rational.

If $b = 0$ or $c = 0$, then the matrix is triangular and so the eigenvalues are just the diagonal entries $a^2, d^2$ (which in particular are rational). So, any of the $12$ choices with $\{a, b, c, d\} = \{0, 1, 2, 3\}$ and either $b = 0$ or $c = 0$ gives a solution, and these examples evidently minimize $\max\{a, b, c, d\}$ among nonnegative solutions.

If we exclude $0 \in \{a, b, c, d\}$ to avoid these trivial solutions, the minimal solutions and their eigenvalues are \begin{array}{cc}\hline (a, b, c, d) & \lambda \\ \hline(1, 2, 5, 4) & -4, 21 \\ (4, 2, 3, 5) & 13, 28 \\ \hline \end{array} and the six examples obtained from these using the evident symmetries $a \leftrightarrow d$ and $b \leftrightarrow c$.

For the $3 \times 3$ case already the matrix $$\pmatrix{0^2 & 1^2 & 2^2 \\ 3^2 & 4^2 & 5^2 \\ 6^2 & 7^2 & 8^2}$$ has irrational eigenvalues: Its characteristic polynomial, $c(t) = t^3 - 80 t^2 - 354 t + 216$, has positive discriminant, so it has three real roots. On the other hand, $c(t) \equiv t^3 + t + 1 \pmod 5$, but this latter polynomial has no roots modulo $5$, hence $c(t)$ is irreducible over $\Bbb Q$, that is, its three real roots are irrational.

In a sense that can be made precise, most rational polynomials do not have all roots rational, and I see little reason to expect that the characteristic polynomials of matrices with (distinct) square entries would be special in this regard, so it is perhaps more interesting to ask for an example whose eigenvalues are rational (and hence integral). A quick Maple script (transcribed below) finds many examples with entries $0^2, \ldots, 8^2$. The first of these lexicographically is $$\pmatrix{0^2 & 2^2 & 3^2 \\ 5^2 & 8^2 & 1^2 \\ 7^2 & 6^2 & 4^2} , \quad \textrm{which has eigenvalues} \quad {-13}, 24, 69 .$$ It is rare even among the matrices with entries $0^2, \ldots, 8^2$ to have all rational eigenvalues: It only happens for $252$ of the $9! = 362880$ cases, $180$ of which have no zero eigenvalues.

restart;
with(combinat): with(LinearAlgebra):
m := 3;
N := 9;

for numberSet in choose(N, m^2) do
    shifted := map(U -> U - 1, numberSet);
    print([shifted]);
    for ordering in permute(shifted) do
        map(U -> U^2, ordering);
        A := convert([seq(%[((i - 1) * m + 1)..(i * m)], i=1..m)], Matrix);
        c := CharacteristicPolynomial(A, t);
        if (convert(map(degree, map(U -> U[1], factors(c)[2]), t), set) = {1}) then
            print(ordering, A, solve(c));
        fi:
    od:
od:

Here the constant $m$ is the matrix size, $[0, \ldots, N - 1]$ is the range from which the squared numbers are chosen.