Yes, you can.

Number the columns from left to right and the rows from top to bottom with the numbers $1, 2, \dots, 2022$. Write in the cell $(r, c)$, i.e. the one in the $r$-th row and $c$-th column, the number $x^{2(r+c)}$, where the number $x$ is selected later.

Consider an arbitrary rectangle $R$, which lies in rows with numbers $s, s + 1, \dots, t$ and columns with numbers $u, u + 1, \dots, v$. By construction, the sum of all numbers in the rectangle $R$ is equal to
$$(x^{2s} + x^{2s+2} + \dots + x^{2t}) \cdot (x^{2u} + x^{2u+2} + \dots + x^{2v}) = x^{2s+2u} \cdot \frac{x^{2a} - 1}{x^2 - 1} \cdot \frac{x^{2b} - 1}{x^2 - 1},$$
where $a = t - s + 1$ and $b = v - u + 1$. It is immediately clear that if $a = b$, the sum of numbers in $R$ is the perfect square of any real number $x$. Now it remains to find such an $x$ that no matter how $a \neq b$, where $a, b \in \{2, 3, \dots, 2021\}$, the number $(x^{2a} - 1)(x^{2b} - 1)$ would not be a perfect square.

Proving. Let $P(x) = x^{2n} + a_{2n-1}x^{2n-1} + \dots + a_1x + a_0$. We can immediately see that starting from some point $P(x)$ is positive, because its highest coefficient is $1$. In the following, we will consider only $x$ from this point onwards. First, let's find the following polynomial $Q(x) = x^n + b_{n-1}x^{n-1} + \dots + b_1x + b_0$ with rational coefficients such that the degree of the polynomial $P(x) - Q(x)^2$ is less than $n$. To do this, it is enough to restore the coefficients one by one $b_{n-1}, \dots, b_1, b_0$, using identities:
$$\sum_{k=n}^{2n-1} \sum_{i+j=k} b_i b_j x^k = a_k x^k.$$
Thus, we can write the equality $P(x) = Q(x)^2 + R(x)$, $\deg R < n$. Let's multiply both parts of this equality by $M^2$ so that the polynomials $M^2P(x) = P_1(x)$, $MQ(x) = H(x)$ and $M^2R(x) = S(x)$ become integer coefficients. We have that $P_1(x) = H^2(x) + S(x)$, where again $\deg S < n$ and all polynomials $P_1, H$ and $S$ have integer coefficients. Suppose that what you want to prove is not true. This would mean that $S(x)$ is not identically zero. We know that there are infinitely many $x$ such that $P_1(x) = M^2P(x)$ is a perfect square. We will see that since the coefficients of the polynomial $S(x)$ are integer and nonnegative and $S(x)$ is not exactly zero, starting from some point $S(x) \neq 0$, and therefore from the same point $H(x)^2 \neq P_1(x)$. Then consider sufficiently large positive integers $t$, of which there are infinitely many, such that $P_1(t) = w^2$, where $w$ is a positive integer and $H(t)^2 \neq P_1(t)$. We see
$$|P_1(t) - H^2(t)| \geq 2w - 1 \Rightarrow |S(t)| \geq 2w - 1 = 2\sqrt{P(t)} - 1 \geq 2t^n - 1 > t^n.$$
But since $\deg S < n$, the inequality $S(t) > t^n$ can only hold for a finite number of different $t$, which leads to a contradiction.

The lemma implies that for any given $a$ and $b$, there are only a finite number of $x$ such that $(x^{2a} - 1)(x^{2b} - 1)$ is a perfect square. Indeed, otherwise, there would be a polynomial $Q(x)$ with rational coefficients such that $(x^{2a} - 1)(x^{2b} - 1) = Q^2(x)$. But we can see that there is a complex root $e^{\frac{2\pi i}{a}}$ on the left-hand side, which has a multiplicity of $1$, and this is where we get the contradiction. Alternatively, we can see that $(x^{2a} - 1)' = 2a x^{2a-1}$, which has a GCD of $1$ with $(x^{2a} - 1)$, and therefore $x^{2a} - 1$ has no roots of multiplicity. Therefore, for any given $a$ and $b$, there are only a finite number of $x$ such that $(x^{2a} - 1)(x^{2b} - 1)$ is a perfect square. And since there are a finite number of ways to choose a pair $(a, b)$, it means that there is an $x_0$, for which $(x_0^{2a} - 1)(x_0^{2b} - 1)$ is not a perfect square for any positive integers $a$ and $b$ not greater than $2019$. Therefore, we can put $x = x_0$ and the statement is proved.