Does $A\times A\cong B\times B$ imply $A\cong B$? This is similar to What does it take to divide by $2$? about $(A\sqcup A\cong B\sqcup B)\Rightarrow A\cong B$ which is valid in $\textsf{ZFC}$ by using cardinalities and also in $\textsf{ZF}$ by some combinatorial argument, but where it remained unclear whether constructive arguments suffice.
This questions concerns products instead of disjoint union, so:

Question: Does $\textsf{ZF}$ prove the implication $(A\times A\cong B\times B)\Rightarrow A\cong B$?

If yes: is anything known in case of intuitionistic set/type theories?
 A: This is a generalization of tetori's nice answer, which in fact shows that this assumption implies the axiom of choice.

Lemma I. If $a$ and $b$ are infinite cardinals equipollent with their own squares, then $(a+b)^2=a\cdot b$.

Proof. We calculate:
$$(a+b)^2=a^2+b^2+2\cdot a\cdot b=a^2+b^2+a\cdot b=a+b+a\cdot b=a\cdot b$$
The second equality comes from the fact that $a\leq a+a=2\cdot a\leq a^2=a$, the last one follows from the same reasoning. $\qquad\square$

Lemma II. If $a+a=a$ then $2^a$ equals to its own square.

Proof. $(2^a)^2=2^{a+a}=2^a.\qquad\square$

Tarski's Lemma. If $a$ is a cardinal, $\lambda$ is an aleph, and $a+\lambda=a\cdot\lambda$, then the two are comparable.

(The proof appears in Jech, The Axiom of Choice, Lemma 11.6; and there are generalizations of this lemma out there.)

Theorem. ($\sf ZF$) If $a^2=b^2\implies a=b$ for infinite cardinals, then the axiom of choice holds.

Proof. We will show that if $\kappa$ is an aleph number, then $2^\kappa$ is an aleph number. This implies the axiom of choice, by a theorem of Rubin and Rubin from 1963.
Let $\lambda=\aleph(2^\kappa)$, the least aleph cardinal such that $\lambda\nleq 2^\kappa$. Then $2^\kappa$ and $\lambda$ satisfy the hypothesis of Lemma I, and therefore $(2^\kappa+\lambda)^2=2^\kappa\cdot\lambda=(2^\kappa)^2\cdot\lambda^2=(2^\kappa\cdot\lambda)^2$. By the uniqueness of the square root, $2^\kappa+\lambda=2^\kappa\cdot\lambda$, and by Tarski's lemma $2^\kappa$ and $\lambda$ are comparable.
Finally $\lambda\nleq2^\kappa$, therefore $2^\kappa<\lambda$, so $2^\kappa$ is an aleph number. $\qquad\square$
(I will remark here that the final proof uses $\sf ZF$ in a substantial way. In $\sf ZFA$, where atoms are allowed, having $2^\kappa$ as an aleph does not imply the axiom of choice.)
A: This does not answer the question (which I misread, being in a mental fog), but I am undeleting it at the OP's request.
The following summary is copied from p. 174 of Wacław Sierpiński, Cardinal and Ordinal Numbers, second edition revised, Warszawa, 1965:

If, for a natural number $k$ and cardinal numbers $m$ and $n$, we have $km=kn$, then $m=n$.

This theorem was proved without the aid of the axiom of choice for $k=2$ in 1905 by F. Bernstein, who also outlined a proof for natural $k$ (Bernstein [1], p. 122; also Sierpiński [8], p. 1). A detailed proof for natural $k$ (without the aid of the axiom of choice) was published by A. Tarski [12] (p. 77). who proved also (without the aid of the axiom of choice) that if for a natural number $k$ and cardinal numbers $m$ and $n$ we have $km\le kn$, then $m\le n$ (for $k=2$ see also Sierpiński [31], p. 148).
F. Bernstein, Untersuchungen aus der Mengenlehre, Math. Annalen 61 (1905), p. 117-155.
W. Sierpiński, Sur l'égalité $2m=2n$ pour les nombres cardinaux, Fund. Math. 3 (1922), p. 1-16.
W. Sierpiński, Sur l'implication $(2m\le2n)\rightarrow(m\le n)$ pour les nombres cardinaux, Fund. Math. 34 (1947), p. 148-154.
A. Tarski, Cancellation laws in the arithmetic of cardinals, Fund. Math. 36 (1949), p. 77-92.
A: The answer is no. It is known that ZF+'$\aleph_1$ and $2^{\aleph_0}$ are incomparable' is consistent if ZF does. Assume that $\aleph_1$ and $2^{\aleph_0}$ are incomparable (over ZF),then 


*

*$\aleph_1$, $2^{\aleph_0}<\aleph_1+2^{\aleph_0}$ and

*$\aleph_1+2^{\aleph_0}\neq\aleph_1\cdot 2^{\aleph_0}$.


1 is trivial. To prove 2, we can use following theorem:

Theorem (ZF) If $\mathfrak{p}$ is a (possibly non-well-orderable) cardinal and $\alpha$ is an aleph and they satisfy $\mathfrak{p}+\alpha=\mathfrak{p}\cdot\alpha$, then they are comparable.

You can find this theorem and its proof in Jech's 'Axiom of choice', Lemma 11.16. 
We will check that $(\aleph_1+2^{\aleph_0})^2=(\aleph_1\cdot 2^{\aleph_0})^2=\aleph_1\cdot 2^{\aleph_0}$. It is just a calculation:
$$
\begin{align}(\aleph_1+2^{\aleph_0})^2
&=\aleph_1+2^{\aleph_0}+2\cdot \aleph_1\cdot2^{\aleph_0}\\
&=\aleph_1+2^{\aleph_0}+\aleph_1\cdot2^{\aleph_0}\\
&=\aleph_1+(\aleph_1+1)\cdot 2^{\aleph_0}\\
&=\aleph_1+\aleph_1\cdot 2^{\aleph_0}\\
&=\aleph_1\cdot(1+2^{\aleph_0})\\
&=\aleph_1\cdot2^{\aleph_0}.\\
\end{align}$$
Also, the square of $\aleph_1\cdot 2^{\aleph_0}$ is $\aleph_1\cdot 2^{\aleph_0}$ itself so $(\aleph_1+2^{\aleph_0})^2=(\aleph_1\cdot 2^{\aleph_0})^2$. However, we already know that $\aleph_1+2^{\aleph_0}$ and $\aleph_1\cdot 2^{\aleph_0}$ are not equal.
