Lemma 2. The following statement and its converse are both true: If $q, r, s$ are three distinct positive real numbers such that $q, r, 2s$ are side lengths of a triangle, then there is an unique triangle $PQR$ with $PQ = q, PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$.
Proof. If $q, r, 2s$ are side lengths of a triangle, we construct the unique triangle $PRQ_1$ with $PR = r, RQ_1 = q$, and $PQ_1 = 2s$. Let $S$ be the midpoint of side $PQ_1$. Expend segment $RS$ through $S$ to $Q$ so that $PQQ_1R$ is a parallelogram. It is clear that $PQR$ is a triangle with $PQ = q, PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. To prove the converse statement, we only need to note that the above procedure can be reversed. □

Lemma 3. Let $PQR$ be a triangle with $PQ = q, PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. Then the area of triangle $PQR$ is
$$\frac{1}{4}\sqrt{2q^2r^2 + 8r^2s^2 + 8s^2q^2 - q^4 - r^4 - 16s^4}.$$
Proof. Because $PQQ_1R$ is a parallelogram, triangles $PQR$ and $PRQ_1$ have the same area. Because $(PR, RQ_1, Q_1P) = (r, q, 2s)$, the desired results follows directly from the Heron's formula. □

In view of Lemma 2, if triangles $ABC$ and $XYZ$ are a pair of pals, we may assume without of generality that $(AB, AC, AM) = (n, s, t)$ and $(XY, XZ, XW) = (n, t, s)$. By Lemma 3, we have
$$2n^2s^2 + 8s^2t^2 + 8t^2n^2 - n^4 - s^4 - 16t^4 = 2n^2t^2 + 8t^2s^2 + 8s^2n^2 - n^4 - t^4 - 16s^4$$
which simplifies to
$$6n^2t^2 - 6n^2s^2 = 15t^4 - 15s^4.$$
Because $ABC$ and $XYZ$ are incongruent, we deduce that $2n^2 = 5(t^2 + s^2)$. We set $n = 5(k^2 + (k+1)^2) = 10k^2 + 10k + 1$ for some positive integer $k$. By applying the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 - (ad - bc)^2$ repeatedly, we have
$$\begin{aligned} t^2 + s^2 &= 10n^2 = (1^2 + 3^2)(k^2 + (k+1)^2)(k^2 + (k+1)^2) \\ &= ((4k+3)^2 + (2k-1)^2)(k^2 + (k+1)^2) \\ &= (6k^2 + 4k - 1)^2 + (2k^2 + 8k + 3)^2. \end{aligned}$$
We set $(n, s, t) = (10k^2 + 10k + 1, 6k^2 + 4k - 1, 2k^2 + 8k + 3)$ for positive integer $k$. For large $k$, it is easy to see that each of $(n, 2s, t)$ and $(n, s, 2t)$ is a set of side lengths of a triangle. Furthermore, by the Euclidean algorithm, it is not difficult to check these numbers are pairwise relatively prime to each other, hence $(n, 2s, t)$ and $(n, s, 2t)$ are the side lengths of a pair of pals, completing the solution.

The answer is yes.
We start with the following observations.

Lemma 2. The following statement and its converse are both true: If $q, r, s$ are three distinct positive real numbers such that $q, r, 2s$ are side lengths of a triangle, then there is an unique triangle $PQR$ with $PQ = q, PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$.

Proof. If $q, r, 2s$ are side lengths of a triangle, we construct the unique triangle $PRQ_1$ with $PR = r, RQ_1 = q$, and $PQ_1 = 2s$. Let $S$ be the midpoint of side $PQ_1$. Expend segment $RS$ through $S$ to $Q$ so that $PQQ_1R$ is a parallelogram. It is clear that $PQR$ is a triangle with $PQ = q, PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. To prove the converse statement, we only need to note that the above procedure can be reversed. □

Lemma 3. Let $PQR$ be a triangle with $PQ = q, PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. Then the area of triangle $PQR$ is
$$\frac{1}{4}\sqrt{2q^2r^2 + 8r^2s^2 + 8s^2q^2 - q^4 - r^4 - 16s^4}.$$
Proof. Because $PQQ_1R$ is a parallelogram, triangles $PQR$ and $PRQ_1$ have the same area. Because $(PR, RQ_1, Q_1P) = (r, q, 2s)$, the desired results follows directly from the Heron's formula. □

In view of Lemma 2, if triangles $ABC$ and $XYZ$ are a pair of pals, we may assume without loss of generality that $(AB, AC, AM) = (n, s, t)$ and $(XY, XZ, XW) = (n, t, s)$. By Lemma 3, we have
$$2n^2s^2 + 8s^2t^2 + 8t^2n^2 - n^4 - s^4 - 16t^4 = 2n^2t^2 + 8t^2s^2 + 8s^2n^2 - n^4 - t^4 - 16s^4$$
which simplifies to
$$6n^2t^2 - 6n^2s^2 = 15t^4 - 15s^4.$$
Because $ABC$ and $XYZ$ are incongruent, we deduce that $2n^2 = 5(t^2 + s^2)$. We set $n = 5(k^2 + (k+1)^2) = 10k^2 + 10k + 1$ for some positive integer $k$. By applying the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 - (ad - bc)^2$ repeatedly, we have
$$\begin{aligned} t^2 + s^2 &= 10n^2 = (1^2 + 3^2)(k^2 + (k+1)^2)(k^2 + (k+1)^2) \\ &= ((4k+3)^2 + (2k-1)^2)(k^2 + (k+1)^2) \\ &= (6k^2 + 4k - 1)^2 + (2k^2 + 8k + 3)^2. \end{aligned}$$
We set $(n, s, t) = (10k^2 + 10k + 1, 6k^2 + 4k - 1, 2k^2 + 8k + 3)$ for positive integer $k$. For large $k$, it is easy to see that each of $(n, 2s, t)$ and $(n, s, 2t)$ is a set of side lengths of a triangle. Furthermore, by the Euclidean algorithm, it is not difficult to check these numbers are pairwise relatively prime to each other, hence $(n, 2s, t)$ and $(n, s, 2t)$ are the side lengths of a pair of pals, completing the solution.